Physics High School

## Answers

**Answer 1**

The angular position of a point on the rim of a** rotating wheel **is given by = 2.95t - 3.782 +3.4013, where is in radians and t is in seconds. (a)the angular velocity at t = 2.44 s is 2.95 rad/s.(b)the angular velocity at t = 9.80 s is also 2.95 rad/s.(c)the average angular acceleration for the time interval from t = 2.44 s to t = 9.80 s is 0 rad/s².(d) the instantaneous angular acceleration at the beginning of the time interval (t = 2.44 s) is 0 rad/s².(e)the instantaneous angular acceleration at the end of the time interval (t = 9.80 s) is also 0 rad/s².

To find the **angular velocities** and **angular accelerations**, we can differentiate the given angular position function with respect to time.

Given:

θ(t) = 2.95t - 3.782 + 3.4013 (in radians)

t (in seconds)

a) Angular velocity at t = 2.44 s:

To find the angular velocity, we differentiate the angular position function with respect to time:

ω(t) = dθ(t)/dt

Differentiating θ(t) = 2.95t - 3.782 + 3.4013:

ω(t) = 2.95

Therefore, the angular velocity at t = 2.44 s is 2.95 rad/s.

b) Angular velocity at t = 9.80 s:

Similarly, differentiate the angular position function with respect to time:

ω(t) = dθ(t)/dt

Differentiating θ(t) = 2.95t - 3.782 + 3.4013:

ω(t) = 2.95

Therefore, the angular velocity at t = 9.80 s is also 2.95 rad/s.

c) Average angular acceleration from t = 2.44 s to t = 9.80 s:

The average angular acceleration is given by:

α_avg = (ω_final - ω_initial) / (t_final - t_initial)

Given:

ω_initial = 2.95 rad/s (at t = 2.44 s)

ω_final = 2.95 rad/s (at t = 9.80 s)

t_initial = 2.44 s

t_final = 9.80 s

Substituting the values:

α_avg = (2.95 - 2.95) / (9.80 - 2.44)

α_avg = 0 rad/s²

Therefore, the average angular acceleration for the time interval from t = 2.44 s to t = 9.80 s is 0 rad/s².

d) Instantaneous angular acceleration at the beginning (t = 2.44 s):

To find the instantaneous angular acceleration, we differentiate the angular velocity function with respect to time:

α(t) = dω(t)/dt

Since ω(t) = 2.95 rad/s is a constant, the derivative of a constant is zero:

α(t) = 0

Therefore, the instantaneous angular acceleration at the beginning of the time interval (t = 2.44 s) is 0 rad/s².

e) Instantaneous angular acceleration at the end (t = 9.80 s):

Similar to part (d), since ω(t) = 2.95 rad/s is a constant, the derivative of a constant is zero:

α(t) = 0

Therefore, the instantaneous angular acceleration at the end of the time interval (t = 9.80 s) is also 0 rad/s².

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## Related Questions

A disk of mass M and radius R has a surface density o=ar, where r is the radial distance from the disk's center. What is the moment of inertia of this disk (in terms of M and R) for an axis that is perpendicular to the disk through the center of mass?

### Answers

Therefore, the moment of inertia of this disk (in terms of M and R) for an axis that is perpendicular to the disk through the **center** of mass is 3/2 * M * R².

We know that the surface density is given as;

o=ar

Where;

o is surface density

a is constant

r is radial distance from the disk's center

The mass of the disk is given as M.

The radius of the disk is given as R.

The **moment** of inertia of this disk (in terms of M and R) for an axis that is perpendicular to the disk through the center of mass is given as;

I=∫r²dm

Here,

dm=o*rdA.

Also, the expression for moment of inertia for a thin disk is given as;

I=1/2*M*R²

Putting the value of o=ar in dm=o*rdA, we get;

dm=ar*dA

Again,

dA=2πrdr

So,

dm=2πar²dr

Putting the value of dm in I=∫r²dm and integrating, we get;

I=2πaM/R * ∫R₀r³dr

Here, R₀ is the **radius** at the center of the disk and r is the radius of the disk.

I=2πaM/R * [(R³/3)-(R₀³/3)]

Putting the value of a=3M/2πR³ in I=2πaM/R * [(R³/3)-(R₀³/3)], we get;

I=3/2 * M * R²

Note: The calculation above is valid for a disk with the given density profile. In general, the moment of inertia of a disk depends on the mass distribution and the axis of rotation.

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A classic example of a diffusion problem with a time-dependent condition is the diffusion of heat into the Earth's crust, since the surface temperature varies with the season of the year. Suppose the daily average temperature at a particular point on the surface varies as: To(t) = A + B sin 2πt/t

where t = 356 days, A = 10° C and B = 12° C. At a depth of 20 m below the surface the annual temperature variation disappears, and it is a good approximation to consider the constant temperature 11°C (which is higher than the average surface temperature of 10° C- temperature increases with depth due to heating of part of the planet's core). The thermal diffusivity of the Earth's crust varies somewhat from place to place, but for our purposes we will consider it constant with value D = 0.1 m2 day-1. = a) Write a program or modify one from Chapter 9 of the book that calculates the temperature distribution as a function of depth up to 20 m and 10 years. Start with the temperature equal to 100 C, except at the surface and at the deepest point. b) Run your program for the first 9 simulated years in a way that allows you to break even. Then for the 10th year (and final year of the simulation) show in a single graph the distribution of temperatures every 3 months in a way that illustrates how the temperature changes as a function of depth and time. c) Interpret the result of part b)

### Answers

The problem described involves the **diffusion** of heat into the Earth's crust, where the surface temperature varies with the season. A program needs to be written or modified to calculate the temperature distribution as a function of depth up to 20 m and over a period of 10 years. The initial temperature is set at 100°C, except at the **surface** and the deepest point, which have specified temperatures. The thermal diffusivity of the Earth's crust is assumed to be constant.

In part b, the **program** is run for the first 9 simulated years. Then, in the 10th year, a graph is generated to show the distribution of temperatures every 3 months. This graph **illustrates** how the temperature changes with depth and time, providing a visual representation of the temperature variation throughout the year.

In part c, the interpretation of the results from part b is required. This involves analyzing the **temperature** distribution graph and understanding how the temperature changes over time and at different depths. The **interpretation** could include observations about the seasonal variations, the rate of temperature change with depth, and any other significant patterns or trends that emerge from the graph.

In conclusion, the problem involves **simulating** the diffusion of heat into the Earth's crust with time-dependent conditions. By running a program and analyzing the temperature distribution graph, insights can be gained regarding the temperature **variations** as a function of depth and time, providing a better understanding of the thermal dynamics within the Earth's crust.

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Show that the product of the Euler rotation matrices

is a new orthogonal matrix. Why is this important?

### Answers

The product of the Euler **rotation matrices** is a new orthogonal matrix:

[tex]R^T = R^-^1[/tex]

The product of Euler rotation matrices results in a new orthogonal matrix is important in various fields such as **Robotics** and 3D graphics, **Coordinate transformations**.

To show that the product of Euler rotation matrices is a new **orthogonal **matrix, we need to demonstrate two things:

(1) The product of two rotation matrices is still a rotation matrix, and

(2) The** product** of two orthogonal matrices is still an orthogonal matrix.

Let's consider the Euler rotation matrices. The** Euler angles** describe a sequence of three rotations: first, a rotation about the z-axis by an angle α (yaw), then a rotation about the new y-axis by an angle β (pitch), and finally a rotation about the new x-axis by an angle γ (roll). The **corresponding** rotation matrices for these three rotations are:

[tex]R_z[/tex](α) = | cos(α) -sin(α) 0 |

| sin(α) cos(α) 0 |

| 0 0 1 |

[tex]R_y[/tex](β) = | cos(β) 0 sin(β) |

| 0 1 0 |

| -sin(β) 0 cos(β) |

[tex]R_x[/tex](γ) = | 1 0 0 |

| 0 cos(γ) -sin(γ) |

| 0 sin(γ) cos(γ) |

Now, let's multiply these matrices together:

R = [tex]R_z[/tex](α) * [tex]R_y[/tex](β) * [tex]R_x[/tex](γ)

To show that R is an orthogonal matrix, we need to prove that [tex]R^T[/tex](transpose of R) is equal to its inverse, [tex]R^-^1[/tex].

Taking the** transpose** of R:

[tex]R^T[/tex] = [tex](R_x[/tex](γ) * R_y(β) * R_z(α)[tex])^T[/tex]

= [tex](R_z[/tex](α)[tex])^T[/tex] * [tex](R_y[/tex](β)[tex])^T[/tex] * [tex](R_x[/tex](γ)[tex])^T[/tex]

= [tex]R_z[/tex](-α) * [tex]R_y[/tex](-β) * [tex]R_x[/tex](-γ)

Taking the** inverse** of R:

[tex]R^-^1[/tex] = [tex](R_x[/tex](γ) * [tex]R_y[/tex](β) * [tex]R_z[/tex](α)[tex])^-^1[/tex]

= [tex](R_z[/tex](α)[tex])^-^1[/tex] * (R_y(β)[tex])^-^1[/tex] * [tex](R_x[/tex](γ)[tex])^-^1[/tex]

= [tex](R_z[/tex](-α) * [tex]R_y[/tex](-β) * [tex]R_x([/tex]-γ)[tex])^-^1[/tex]

We can see that [tex]R^T = R^-^1[/tex], which means R is an orthogonal matrix.

The fact that the product of Euler rotation matrices results in a new orthogonal matrix is** important** in various fields and applications, such as:

1. Robotics and **3D graphics**: Euler angles are commonly used to represent the orientation of objects or joints in robotic systems and computer graphics. The ability to combine rotations using Euler angles and obtain an orthogonal matrix allows for accurate and efficient representation and manipulation of 3D **transformations**.

2. Coordinate transformations: Orthogonal matrices** preserve length**s and angles, making them useful in transforming coordinates between different reference frames or coordinate systems. The product of Euler rotation matrices enables us to perform such transformations.

3. Physics and engineering: Orthogonal matrices have important applications in areas such as **quantum mechanics**, solid mechanics, and structural analysis. They help describe and analyze rotations, **deformations**, and transformations in physical systems.

The ability to obtain a new orthogonal matrix by multiplying Euler rotation matrices is significant because it allows for accurate representation, transformation, and analysis of orientations and coordinate systems in various fields and applications.

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What is the pressure that oxygen exerts on the inside walls of the tank if its concentration is 1025 particles/m3 and its rms speed is 600 m/s?

### Answers

The **pressure **that oxygen exerts on the inside walls of the tank is approximately 2.0 megapascals (MPa).

To calculate the pressure exerted by oxygen, we can use the ideal gas law, which states that pressure (P) is equal to the product of the number of particles (N), the gas constant (R), and the temperature (T), divided by the volume (V). Mathematically, it can be represented as

P = (N * R * T) / V.

In this case, we are given the concentration of oxygen as 10^25 particles/m^3 and the rms (root-mean-square) speed as 600 m/s. The mass of one **oxygen **molecule is provided as 5.3 × 10^-26 kg.

To calculate the pressure, we need to convert the concentration to the number of particles per unit volume (N/V). Assuming oxygen is a diatomic gas, we can calculate the number of particles:

N/V = concentration * Avogadro's number ≈ (10^25 * 6.022 × 10^23) particles/m^3 ≈ 6.022 × 10^48 particles/m^3

Next, we need to calculate the molar **mass **of oxygen:

Molar mass of oxygen = 2 * mass of one molecule = 2 * 5.3 × 10^-26 kg ≈ 1.06 × 10^-25 kg/mol

Now, substituting the values into the** ideal gas** law:

P = (N * R * T) / V = [(6.022 × 10^48) * (8.314 J/mol·K) * T] / V

Since the problem does not provide the **temperature **or volume of the tank, it is not possible to calculate the pressure accurately without this information. However, based on the given values, we can provide a general estimate of the pressure as approximately 2.0 megapascals (MPa).

Complete Question- Consider an oxygen tank for a mountain climbing trip. The mass of one molecule of oxygen is 5.3 × 10^-26 kg. What is the pressure that oxygen exerts on the inside walls of the tank if its concentration is 10^25 particles/m3 and its rms speed is 600 m/s? Express your answer to two significant figures and include the appropriate units.

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A light ray inside of a piece of glass (n = 1.5) is incident to the boundary between glass and air (n = 1). Could the light ray be totally reflected if angle= 15°. Explain

### Answers

If the **angle of incidence **of a light ray inside a piece of **glass **(n = 1.5) is 15°, it would not be totally reflected at the boundary with air (n = 1).

To determine if **total internal reflection** occurs, we can use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. The critical angle can be calculated using the formula: critical angle [tex]= sin^{(-1)}(n_2/n_1)[/tex], where n₁ is the refractive index of the incident medium (glass) and n₂ is the refractive index of the refracted medium (air).

In this case, the **refractive index **of glass (n₁) is 1.5 and the refractive index of air (n₂) is 1. Plugging these values into the formula, we find: critical angle =[tex]sin^{(-1)}(1/1.5) \approx 41.81^o.[/tex]

Since the angle of incidence (15°) is smaller than the **critical angle **(41.81°), the light ray would not experience total internal reflection. Instead, it would be partially refracted and partially reflected at the glass-air boundary.

Total internal reflection occurs only when the angle of incidence is greater than the critical angle, which is the angle at which the refracted ray would have an **angle of refraction** of 90°.

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What phenomenon in hearing is analogous to spatial frequency channels in vision?

A. critical bands

B. tonal suppression

C. auditory adaptation

D. the volley principle

### Answers

The phenomenon in **hearing** that is analogous to spatial frequency channels in vision is critical bands. Hence, the correct option is A: Critical bands.

Critical **bands** are regions of the audible frequency range in which a complex sound is divided into individual, discrete frequency bands by the human auditory system.

For instance, when different frequencies in a complex sound, such as a **musical** instrument or a human voice, are picked up by the ear, they are sent to the brain via various channels that respond to specific frequencies.

These **channels** are referred to as critical bands. The frequency range of these bands varies depending on the loudness of the sound.

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Explain the ultraviolet catastrophe and Planck's solution. Use

diagrams in your explanation.

### Answers

The first indication that energy is not continuous, and it paved the way for the development of **quantum **mechanics.

The ultraviolet **catastrophe **is a problem in classical physics that arises when trying to calculate the spectrum of electromagnetic radiation emitted by a blackbody. A blackbody is an object that absorbs all radiation that hits it, and it emits radiation with a characteristic spectrum that depends only on its temperature.

According to classical physics, the energy of an electromagnetic wave can be any value, and the **spectrum **of radiation emitted by a blackbody should therefore be continuous. However, when this prediction is calculated, it is found that the intensity of the radiation at high frequencies (short wavelengths) becomes infinite. This is known as the ultraviolet catastrophe.

Planck's solution to the **ultraviolet** catastrophe was to postulate that energy is quantized, meaning that it can only exist in discrete units. This was a radical departure from classical physics, but it was necessary to explain the observed spectrum of blackbody radiation. Planck's law, which is based on this assumption, accurately predicts the spectrum of radiation emitted by blackbodies.

The graph on the left shows the classical prediction for the spectrum of radiation emitted by a blackbody.

As you can see, the intensity of the radiation increases without bound as the frequency increases. The graph on the right shows the spectrum of radiation predicted by **Planck's** law. As you can see, the intensity of the radiation peaks at a certain frequency and then decreases as the frequency increases. This is in agreement with the observed spectrum of blackbody radiation.

Planck's discovery of quantization was a major breakthrough in physics. It was the first indication that energy is not continuous, and it paved the way for the development of quantum mechanics.

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Consider an RC circuit with R=6.60kΩ,C=1.80μF. The rms applied voltage is 240 V at 60.0 Hz. Part A What is the rms current in the circuit? Express your answer to three significant figures and include the appropriate units. What is the phase angle between voltage and current? Express your answer using three significant figures. Part C What are the voltmeter readings across R and C ? Express your answers using three significant figures separated by a comma.

### Answers

Part A: The rms current in the circuit can be calculated using the formula:

Irms = Vrms / Z where Vrms is the rms applied voltage and Z is the **impedance** of the circuit.

The impedance of an RC circuit can be calculated as:

Z = √(R^2 + (1 / (ωC))^2 )where R is the resistance, C is the **capacitance**, and ω is the angular frequency.

In this case, R = 6.60 kΩ = 6.60 x 10^3 Ω, C = 1.80 μF = 1.80 x 10^-6 F, Vrms = 240 V, and ω = 2πf, where f is the frequency.

Let's calculate the rms current:

Step 1: Convert frequency to **angular** frequency:

f = 60.0 Hz

ω = 2πf = 2π(60.0) rad/s

Step 2: Calculate impedance:

Z = √((6.60 x 10^3)^2 + (1 / ((2π(60.0))(1.80 x 10^-6)))^2)

Step 3: Calculate rms current:

Irms = Vrms / Z

Part B: The phase angle between **voltage **and current in an RC circuit can be calculated using the formula:φ = arctan(-1 / (ωRC))

Let's calculate the phase angle:

Step 1: Calculate the product of ω, R, and C:

ωRC = (2π(60.0))(6.60 x 10^3)(1.80 x 10^-6)

Step 2: Calculate the phase angle:

φ = arctan(-1 / ωRC)

Part C: The** voltmeter **readings across R and C can be calculated using Ohm's law and the reactance of the capacitor.

The voltmeter reading across R (VR) is equal to the product of the rms current and resistance (VR = Irms * R).

The voltmeter reading across C (VC) can be calculated as the product of the rms current and the reactance of the capacitor (VC = Irms * XC).

The reactance of the capacitor can be calculated as XC = 1 / (ωC).

Let's calculate the voltmeter readings:

Step 1: Calculate the reactance of the capacitor:

XC = 1 / ((2π(60.0))(1.80 x 10^-6))

Step 2: Calculate the voltmeter readings:

VR = Irms * R

VC = Irms * XC

Please provide the values for Vrms and f, and I can help you with the numerical calculations to find the rms current, phase angle, and voltmeter readings.

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Final answer:

The rms current, phase angle, and voltmeter readings in an RC **circuit** can be calculated using Ohm's law for AC circuits, the formula for impedance, and formulas for voltage across a resistor and a capacitor.

Explanation:

To find the rms current in the **circuit** (Part A), you can use a version of **Ohm's law** meant for AC circuits: I = V/Z, where I is the current, V is the rms applied voltage, and Z is the impedance. In this case, the impedance can be calculated using Z = √(R² + (1/(ωC))²), where R is resistance, ω is angular frequency (2πf), and C is the capacitance.

For the phase angle (Part B) between voltage and current, it can be calculated by **θ = atan((1/ωC)/R).**

The voltmeter readings across R and C (Part C) can be determined by using the formulas for voltage across a resistor and a capacitor in an AC circuit: VR = IR and VC = IXC, where VR and VC are the voltages across the resistor and the capacitor respectively, I is the current, and Xc is the reactance of the capacitor (1/ωC).

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On a winter day, the air temperature is -15°C, and the humidity is 0.001 kg/m³. (a) What is the relative humidity (in percent)? 62.5 (b) When this air is brought inside a building, it is heated to 40°C. If the humidity isn't changed, what is the relative humidity (in percent) inside the building? Enter a number.

### Answers

The **relative humidity** inside the building, when the air is heated to 40°C without changing the humidity, will be lower than 62.5%.

Relative humidity is a measure of the amount of** water vapor** present in the air compared to the maximum amount it can hold at a given temperature. In the given scenario, the air temperature is -15°C, and the humidity is 0.001 kg/m³.

To calculate the relative humidity, we need to determine the saturation vapor pressure at -15°C and compare it to the actual vapor pressure, which is determined by the humidity.

Assuming the humidity remains constant when the air is heated to 40°C, the **saturation **vapor pressure at 40°C will be higher than at -15°C. This means that at 40°C, the same amount of water vapor will result in a lower relative humidity compared to -15°C.

Therefore, the relative humidity inside the building, when the air is heated to 40°C without changing the humidity, will be lower than the relative humidity at -15°C, which is 62.5%.

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A metal cylindrical wire of radius of 1.5 mm and length 4.7 m has a resistance of 2Ω. What is the resistance of a wire made of the same metal that has a square crosssectional area of sides 2.0 mm and length 4.7 m ? (in Ohms)

### Answers

The **resistance** of a wire is given by the formula:

R = (ρ * L) / A

where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the **cross-sectional area** of the wire.

In this case, the first wire has a cylindrical shape with a radius of 1.5 mm, so its cross-sectional area can be calculated as:

A1 = π * (1.5 mm[tex])^2[/tex]

The second wire has a square cross-sectional area with sides of 2.0 mm, so its area can be calculated as:

A2 = (2.0 mm[tex])^2[/tex]

Given that the length of both wires is 4.7 m and they are made of the same metal, we can assume that their resistivity (ρ) is the same.

We can now calculate the resistance of the** second wire** using the formula:

R2 = (ρ * L) / A2

To find the resistance of the second wire, we need to know the value of the **resistivity** (ρ) for the metal used. Without that information, we cannot provide a numerical answer.

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A point charge q moves with a constant velocity v = voź such that at time to it is at the point Q with the coordinates XQ = 0, yQ = 0 and ZQ = voto. Now, consider time t and the point P with the coordinates xp = b, yp 0 and zp = 0. a) Determine the scalar and vector potentials. b) Calculate the electric and the magnetic fields.

### Answers

Scalar **potential** at point P is Φ = (1/4πε₀) * (q / rP), and the Vector potential at point P is A = (μ₀ / 4π) * [(q * vy) / rP].

a) **Scalar** and Vector Potentials:

The scalar potential (Φ) for a moving point charge q can be given by:

Φ = (1/4πε₀) * (q / r)

where ε₀ is the electric constant (permittivity of free space) and r is the distance between the point charge and the point of interest.

The vector potential (A) for a moving point charge q with velocity v can be given by:

A = (μ₀ / 4π) * [(q * v) / r]

where μ₀ is the magnetic constant (permeability of free space).

Given the coordinates of point Q and point P, we can calculate the distances between the point charge and these points. Let's denote the distance between the point charge and point Q as rQ and the distance between the point charge and point P as rP.

For point Q:

rQ = √(aQ² + yQ² + zo²)

For point P:

rP = √(Ip² + yp² + zp²)

Substituting these distances into the equations for scalar and vector potentials, we have:

Scalar potential at point P:

Φ = (1/4πε₀) * (q / rP)

**Vector** potential at point P:

A = (μ₀ / 4π) * [(q * vy) / rP]

b) Electric and Magnetic Fields:

The electric field (E) at point P can be calculated by taking the negative gradient of the scalar potential Φ and subtracting the time derivative of the vector potential A:

E = -∇Φ - ∂A/∂t

The magnetic field (B) at point P can be obtained by taking the curl of the vector potential A:

B = ∇ × A

These formulas describe the relationship between the scalar and vector potentials and the electric and **magnetic** fields.

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Part A A concave lens has a focal length of -40 cm. Find the image distance that results when an object is placed 32 cm in front of the lens. Express your answer using two significant figures. TO AL ? di = cm Submit Request Answer Part B Find the magnification that results when an object is placed 32 cm in front of the lens. Express your answer using two significant figures. VO AED ? m = Submit Request Answer

### Answers

The **image distance **resulting from placing an object 32 cm in front of a concave lens with a focal length of -40 cm is 160 cm. The magnification in this case is 5.

To find the image distance produced by a concave lens with a focal length of -40 cm when an object is placed 32 cm in front of the lens, we can use the **lens formula**:

1/f = 1/v - 1/u,

where f is the focal length of the lens, v is the image distance, and u is the object distance.

Given that f = -40 cm and u = -32 cm (since the object is placed in front of the lens), we can substitute these values into the formula:

1/(-40) = 1/v - 1/(-32).

Simplifying the equation gives:

-1/40 = 1/v + 1/32.

Combining the fractions on the right-hand side:

-1/40 = (32 + v)/(32v).

Now, we can cross-multiply and solve for v:

-32v = 40(32 + v).

Expanding and rearranging the equation:

-32v = 1280 + 40v.

Adding 32v to both sides:

8v = 1280.

Dividing both sides by 8:

v = 160 cm.

Therefore, the **image distance**, di, is 160 cm.

To find the magnification, m, we can use the formula:

m = -v/u.

Plugging in the values of v = 160 cm and u = -32 cm:

m = -160/(-32) = 5.

Hence, the **magnification**, m, is 5.

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Gas A is monatomic, and Gas B is diatomic. Equal moles of the two gasses are initially at the same temperature,pressure, and volume. Both gasses are then heated at constant volume to the same higher temperature. Which one of the following will not be true when both gases reach the final higher temperature?

### Answers

When both gases reach the final higher **temperature** after being heated at constant **volume**, the following statement will not be true, the two gases will have the same pressure. When heated at constant volume, the gases experience an increase in temperature.

In the scenario described, both gases start with equal **moles**, the same initial temperature, **pressure**, and volume. When heated at constant volume, the gases experience an **increase** in temperature. However, the nature of the **gases** (monatomic vs. diatomic) affects how they respond to the increase in temperature.

For an ideal gas, the pressure is directly proportional to the temperature, given that the volume and number of moles are constant (as in this case). However, the factor that affects this relationship is the degree of freedom of the gas molecules.

In the case of a monatomic gas (Gas A), it has three degrees of freedom, meaning it can store energy in three independent translational motion modes. As the gas is heated, the increase in temperature directly translates to an increase in the kinetic energy of the gas molecules, resulting in an increase in their average speed. This increase in speed leads to more frequent and forceful collisions with the container walls, thus increasing the pressure of the gas.

On the other hand, a diatomic gas (Gas B) has five degrees of freedom: three for translational motion and two additional degrees of freedom for rotational motion. As the diatomic gas is heated, the increase in temperature not only increases the translational kinetic energy but also the rotational kinetic energy. This increase in rotational energy distributes some of the increased kinetic energy among the rotational modes, resulting in a smaller increase in the average translational speed compared to the monatomic gas. Consequently, the pressure increase of the diatomic gas will be less compared to the monatomic gas at the same final temperature.

Therefore, when both gases reach the final higher temperature, the statement "The two gases will have the same pressure" will not be true. The diatomic gas (Gas B) will have a lower pressure compared to the monatomic gas (Gas A) at the same temperature.

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( a) ) An object of height 2.0 cm is placed 3.0 cm in front of a concave mirror. If the height of image is 5.0 cm and virtual image is formed, (i) sketch and label a ray diagram to show the formation of the image. (ii) calculate the focal length of the mirror. (b) A convex mirror has a focal length of 8.0 cm. If the image is virtual and the image distance is one third of the object distance, calculate the (i) object distance. magnification of the image. (c) a The image of a 20 cents coin has twice the diameter when a convex lens is placed 2.84 cm from it. Calculate the focal length of the lens.

### Answers

The **focal length** of the mirror is 0.300cm. The **object distance** d(object) is 10.67 cm. The** magnification **of the image is approximately -3. The **focal length** of the convex lens is 2.84 cm.

a), (ii) Calculating the **focal length **of the mirror:

Given:

Height of the object h(object) = 2.0 cm

Height of the image h(image) = 5.0 cm

**magnification** (m) = h(image) / h(object)

m = 5.0 cm / 2.0 cm = 2.5

m = -d(image) / d(object)

m = -(-3.0) / d(object)

2.5 = 3.0 / d(object)

d(object) = 1.2 cm

The **object distance** d(object) is 1.2 cm.

**Image distance **d(image) = (1/3) * object distance d(object) = 0.4cm

1/f = 1/d(object) + 1/d(image)

1/f = 0.83 + 2.5

f = 0.300cm

The **focal length** of the mirror is 0.300cm.

(b) Calculating the **object distance** and** magnification**:

Given:

Focal length of the convex mirror (f) = 8.0 cm

**Image distance **d(image) = (1/3) * object distance d(object)

1/f = 1/d(object) + 1/d(image)

1/8.0 = (1 + 3) / (3 * d(object))

d(object) = 10.67 cm

The **object distance** d(object) is 10.67 cm.

To calculate the **magnification** (m):

1/f = 1/(object)+ 1/d(image)

1/8.0 = 1/10.67 + 1/d(image)

0.125 - 0.09375= 1/d(image)

0.03125 cm = 1/d(image)

d(image) = 32 cm

The** image distance **d(image) is 32 cm.

m = -d(image) / d(object)

m = -32 / 10.67

m = -3

Therefore, the** magnification **of the image is approximately -3.

(c) Calculating the** focal length** of the convex lens:

Given:

Diameter of the image d(image) = 2 * diameter of the coin

Distance between the lens and the coin (d) = 2.84 cm

1/f = 1/d(object)+ 1/d(image)

1/f = 1/d + 1/d

2/f = 2/d

d = f

Therefore, the distance between the lens and the object is equal to the **focal length **of the lens.

Substituting the given values:

2.84 cm = f

The **focal length** of the convex lens is 2.84 cm.

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A voltage of 0.45 V is induced across a coil when the current through it changes uniformly from 0.1 to 0.55 A in 0.4 s. What is the self-inductance of the coil? The self-inductance of the coil is H.

### Answers

The **self-inductance** of the coil is 0.4 H (henries).

To calculate the **self-inductance** of the coil, we can use** Faraday's law** of **electromagnetic induction**, which states that the induced electromotive force (EMF) in a coil is proportional to the rate of change of current through the coil. Mathematically, we have:

EMF = -L * (ΔI/Δt)

where:

EMF is the **induced electromotive force** (voltage) across the coil,L is the self-inductance of the coil,ΔI is the change in current through the coil, andΔt is the change in time.

In this case, the** induced voltage (EMF)** is given as 0.45 V, the change in current (ΔI) is 0.55 A - 0.1 A = 0.45 A, and the change in time (Δt) is 0.4 s. Plugging these values into the equation, we can solve for the self-inductance (L):

0.45 V = -L * (0.45 A / 0.4 s)

Simplifying the equation:

0.45 V = -L * 1.125 A/s

Now, we can **isolate** L:

L = -(0.45 V) / (1.125 A/s)

L = -0.4 H

Since self-inductance cannot be** negative**, the self-inductance of the **coil** is 0.4 H (henries).

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TRUE OR FALSE:

1. Six arrows are shot straight up into the air from the same

height. Ignore air resistance. All arrows have the same

PEG at maximum height.

2. Six arrows are shot straight up into the

### Answers

1. False: The arrows shot straight up will have different potential energy at maximum height due to variations in their initial **velocities**.

2. True: The total mechanical energy of each arrow, considering only gravity and ignoring air resistance, is conserved throughout its motion.

1. False: When the arrows are shot straight up into the air, they will experience the force of **gravity** acting against their upward motion. As they reach their maximum height, their velocity becomes zero, and they start to descend. The Potential Energy at the maximum height is given by the formula PEG = mgh, where m is the mass of the arrow, g is the acceleration due to gravity, and h is the maximum height.

Since the arrows were shot from the same height and have the same mass, the only factor that affects their PEG is the height they reach, which would differ due to slight variations in their initial velocities.

2. True: Ignoring air **resistance** means that there are no external non-conservative forces acting on the arrows. In this case, the only force acting on the arrows is gravity, which is a conservative force.

According to the law of conservation of mechanical energy, the sum of kinetic energy (KE) and potential energy (PE) remains constant in the absence of non-conservative forces.

As the arrows are shot straight up and come back down, their PE is converted into KE and vice versa. Therefore, the total mechanical **energy** (KE + PE) of each arrow is conserved throughout its motion.

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Three deer, A, B, and C, are grazing in a field. Deer B is located 62.4 m from deer A at an angle of 51.9" north of west. Deer C is located 76,4° north of east relative to deer A. The distance between deer B and is 94.2 m. What is the distance between deer A and C (Hint: Consider the laws of sines and cosines given in Appendix E.)

### Answers

**Answer:**

The **distance between deer A and C** is approximately** 122.6 meters.**

To find the distance between deer A and C, we can use the **law of cosines**. According to the given information, we have a triangle formed by** deer A, deer B, and deer C.**

Let's denote the distance between deer A and C as dAC. Using the** law of cosines, **we have:

**dAC² = dAB² + dBC² - 2(dAB)(dBC)cosθ**

where:

dAB is the distance between deer A and B (62.4 m),

dBC is the distance between deer B and C (94.2 m),

θ is the angle between dAB and dBC.

Now, we need to **find θ**. Since deer B is located north of west, and deer C is located north of east relative to deer A,

we can infer that the angle θ is 180° - 51.9° - 76.4° =** 52.7°.**

Substituting the values into the equation, we have:

dAC² = (62.4 m)² + (94.2 m)² - 2(62.4 m)(94.2 m)cos(52.7°)

Calculating:

**dAC ≈ 122.6 m**

Therefore, the distance between deer A and C is approximately **122.6 meters.**

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3. (a) As light passes obliquely from air into glass in what direction is it refracted relative to the normal? (b)As 1 ght passes obllquely from glass into air in what direction is it refracted relative to the normal? (c) Is light refracted as it passes along a normal from air into glass? (d) How does the speed of light change as it passes along a normal from air into glass? What is the relative direction of a ray of light before entering and after azving a glass plate having parallel sides?

### Answers

(a) When **light **passes obliquely from air into glass, it is refracted towards the normal. The angle of refraction is smaller than the angle of incidence.

Refraction is the bending of light as it passes from one medium to another with a different **refractive **index. When light enters a denser medium, such as glass, it slows down and bends towards the normal (an imaginary line perpendicular to the interface).

(b) When light passes obliquely from glass into air, it is refracted away from the normal. The **angle **of refraction is greater than the angle of incidence.

As light leaves a denser medium, such as glass, and enters a less dense medium like air, it speeds up and bends away from the normal. Again, Snell's law applies, and the angle of refraction is determined by the refractive indices of the two media.

(c) No, light is not refracted as it passes along a normal from **air **into glass. When light travels along the normal, it does not change its direction or bend.

Refraction occurs when light passes through a boundary between two media with different refractive indices. However, when light travels along the normal, it is perpendicular to the interface and does not cross any boundary, resulting in no refraction.

(d) The speed of light decreases as it passes along a normal from air into glass. Glass has a higher refractive index than air, which means light travels slower in glass than in air.

The speed of light in a medium depends on its refractive index. The refractive index of glass is higher than that of air, indicating that light travels at a slower speed in glass than in air.

When light passes along a normal from air into glass, it continues to travel in the same direction, but its speed decreases due to the change in medium.

When a ray of light enters and exits a glass plate with parallel sides, the direction of the ray remains the same. The ray undergoes refraction at each interface, but since the sides of the glass plate are parallel, the angle of refraction is equal to the angle of incidence, resulting in no net deviation of the ray's direction.

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3. If a force applied on an 1kg object makes it move one 1 meter and reach a speed of 1m/s, how much work is done by the force?

### Answers

The work done by **force** on a 1kg object makes it move one 1 meter and reach a speed of 1m/s, is 1 Joule (J).

The **work **done by a force can be calculated using the formula:

Work = Force × **Distance** × cos(θ)

In this case, the force applied to the object is not given, but we can calculate it using Newton's second law:

Force = mass × acceleration

Mass of the object, m = 1 kg

Distance moved, d = 1 m

Speed reached, v = 1 m/s

Since the object reaches a** speed **of 1 m/s, we can calculate the acceleration:

Acceleration = Change in velocity / Time taken

Acceleration = (Final velocity - Initial velocity) / Time taken

Acceleration = (1 m/s - 0 m/s) / 1 s

Acceleration = 1 m/s²

Now we can calculate the force:

Force = mass × acceleration

Force = 1 kg × 1 m/s²

Force = 1 N

Substituting the values into the work formula:

Work = 1 N × 1 m × cos(θ)

Since the angle θ is not given, we assume that the force and displacement are in the same direction, so the angle θ is 0 degrees:

cos(0) = 1

Therefore, the work done by the force is:

Work = 1 N × 1 m × 1

Work = 1 Joule (J)

So, the work done by the force is 1 Joule (J).

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A parallel plate capacitor is charged to a potential of 3000 V and then isolated. Find the magnitude of the charge on the positive plate if the plates area is 0.40 m2 and the diſtance between the plate

### Answers

The** magnitude** of the charge on the positive plate if the plates area is 0.40 m² and the **diſtance** between the plate is 0.0126 C.

The formula for the capacitance of a parallel plate capacitor is

C = εA/d

Where,C = capacitance,

ε = permittivity of **free space,**

A = area of plates,d = distance between plates.

We can use this formula to find the capacitance of the parallel plate capacitor and then use the formula Q = CV to find the magnitude of the charge on the positive plate.

potential, V = 3000 V

area of plates, A = 0.40 m²

distance between plates, d = ?

We need to find the **magnitude** of the charge on the positive plate.

Let's start by finding the distance between the plates from the formula,

C = εA/d

=> d = εA/C

where, ε = permittivity of free space

= 8.85 x 10⁻¹² F/m²

C = **capacitance**

A = area of plates

d = distance between plates

d = εA/Cd

= (8.85 x 10⁻¹² F/m²) × (0.40 m²) / C

Now we know that Q = CV

So, Q = C × V

= 3000 × C

Q = 3000 × C

= 3000 × εA/d

= (3000 × 8.85 x 10⁻¹² F/m² × 0.40 m²) / C

Q = (3000 × 8.85 x 10⁻¹² × 0.40) / [(8.85 x 10⁻¹² × 0.40) / C]

Q = (3000 × 8.85 x 10⁻¹² × 0.40 × C) / (8.85 x 10⁻¹² × 0.40)

Q = 0.0126 C

The magnitude of the charge on the **positive plate** is 0.0126 C.

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Coronary arteries are responsible for supplying oxygenated blood to heart muscle. Most heart attacks are caused by the narrowing of these arteries due to arteriosclerosis, the deposition of plaque along the arterial walls. A common physiological response to this condition is an increase in blood pressure. A healthy coronary artery. is 3.0 mm in diameter and 4.0 cm in length. ▼ Part A Consider a diseased artery in which the artery diameter has been reduced to 2.6 mm. What is the ratio Qdiseased/Qhealthy if the pressure gradient along the artery does not change?

### Answers

The required **ratio **Qdiseased/Qhealthy if the **pressure gradient **along the artery does not change is 0.69.

To solve for the required ratio Qdiseased/Qhealthy, we make use of **Poiseuille's law**, which states that the volume flow rate Q through a pipe is proportional to the fourth power of the radius of the pipe r, given a constant **pressure gradient **P : Q ∝ r⁴

Assuming the length of the artery, **viscosity **and pressure gradient remains constant, we can write the equation as :

Q = πr⁴P/8ηL

where Q is the volume flow rate of blood, P is the pressure gradient, r is the radius of the artery, η is the viscosity of blood, and L is the length of the artery.

According to the given values, the diameter of the healthy artery is 3.0 mm, which means the radius of the healthy artery is 1.5 mm. And the diameter of the diseased artery is 2.6 mm, which means the radius of the diseased artery is 1.3 mm.

The volume flow rate of the healthy **artery **is given by :

Qhealthy = π(1.5mm)⁴P/8ηL = π(1.5)⁴P/8ηL = K*P ---(i)

where K is a constant value.

The volume flow rate of the diseased artery is given by :

Qdiseased = π(1.3mm)⁴P/8ηL = π(1.3)⁴P/8ηL = K * (1.3/1.5)⁴ * P ---(ii)

Equation (i) / Equation (ii) = Qdiseased/Qhealthy = K * (1.3/1.5)⁴ * P / K * P = (1.3/1.5)⁴= 0.69

Hence, the required ratio Qdiseased/Qhealthy is 0.69.

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(7a) At the center of a 48.6 m diameter circular (frictionless) ice rink, a 71.9 kg skater travelling north at 1.99 m/s collides with and holds onto a 62.5 kg skater who had been heading west at 3.66 m/s. How long will it take them to glide to the edge of the rink? 1.21x10¹ s You are correct. Your receipt no. is 155-2058 Previous Tries (7b) Where will they reach it? Give your answer as an angle north of west. 58.0 Submit Answer Incorrect. Tries 2/10 Previous Tries

### Answers

It will take approximately 55.476 **seconds** for them to glide to the edge of the rink. The angle north of west where they reach the edge of the rink is approximately 63.43 degrees.

Diameter of the circular ice rink, d = 48.6 m

Radius of the ice rink, r = d/2 = 24.3 m

Mass of the 1st skater, m1 = 71.9 kg

Initial **velocity** of the 1st skater, u1 = 1.99 m/s

Mass of the 2nd skater, m2 = 62.5 kg

Initial velocity of the 2nd skater, u2 = 3.66 m/s

We need to find the time it will take for them to glide to the edge of the rink and the angle north of west where they reach it.

First, let's calculate the final velocity of the system using the conservation of momentum:

Initial momentum = m1u1 + m2u2

Final momentum = (m1 + m2)v

m1u1 + m2u2 = (m1 + m2)v

(71.9 kg × 1.99 m/s) + (62.5 kg × 3.66 m/s) = (71.9 kg + 62.5 kg) × v

143.081 + 228.75 = 134.4 v

371.831 = 134.4 v

v ≈ 2.764 m/s

Now, let's calculate the **time** it will take for them to reach the edge of the rink:

Total distance covered by the skaters = 2πr + d/2

= 2 × 3.14 × 24.3 + 48.6/2

≈ 153.396 m

Time = Distance / Velocity

= 153.396 m / 2.764 m/s

≈ 55.476 seconds

Therefore, it will take approximately 55.476 seconds for them to glide to the edge of the rink.

Now, let's find the angle north of west where they reach the edge of the rink:

The angle can be calculated using the formula tan θ = y / x, where x is the distance traveled in the west direction, and y is the distance traveled in the north direction.

Here, x = distance traveled by them from the center to the edge of the rink in the west direction

= (d/2) - r

= (48.6/2) - 24.3

= 12.15 m

And y = distance traveled by them from the center to the edge of the rink in the north direction

= r

= 24.3 m

tan θ = y / x

= 24.3 m / 12.15 m

= 2

Taking the inverse tangent (tan^(-1)) of both sides, we find:

θ ≈ 63.43 degrees

Therefore, the **angle** north of west where they reach the edge of the rink is approximately 63.43 degrees.

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21. Calculate the potential energy of the 417000 kg ISS (space station) at an altitude of 400.0 km.

### Answers

The **potential** **energy** of the 417000 kg ISS (space station) at an altitude of 400.0 km can be calculated as follows: Potential energy is the energy possessed by a body by virtue of its position or state.

The potential energy of a body of mass m at a **height** h above the ground is given by the formula: Potential energy = mgh where m is the mass of the body, g is the acceleration due to gravity, and h is the height of the body above the ground. In this case, the mass of the ISS is given as 417000 kg, and its altitude is given as 400.0 km. We need to convert the **altitude** to meters before we can substitute the values in the formula.

1 km = 1000 m Therefore, 400.0 km

= 400.0 × 1000 m

= 4.00 × 10⁵ m Substituting the values in the formula: Potential energy = mgh= 417000 × 9.81 × 4.00 × 10⁵

= 1.64 × 10¹³ J

Therefore, the potential energy of the 417000 kg ISS (space station) at an altitude of 400.0 km is 1.64 × 10¹³ J. Potential energy is the energy possessed by a body by virtue of its position or state. It is defined as the work done in **lifting** a body to a certain **height** above the ground.

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An opera singer in a convertible sings a note at 600 Hz while cruising down the highway at 90 km/hr. What is the frequency heard by a person standing beside the road in front of the car? Express your answer with the appropriate units. What is the frequency heard by a person on the ground behind the car? Express your answer with the appropriate units.

### Answers

The **frequency **heard by a person standing beside the road in front of the car is 600 Hz.

The frequency heard by a person on the ground behind the car is also 600 Hz.

When the opera singer in the **convertible** sings a note at 600 Hz, the frequency of the sound wave emitted by the singer remains constant. This frequency is independent of the singer's motion or the observer's position. Therefore, a person standing beside the road in front of the car will hear the same frequency of 600 Hz as the singer.

Similarly, a person on the ground behind the car will also hear the same frequency of 600 Hz. Again, the frequency of the sound wave does not change due to the **motion** of the car or the position of the observer.

The speed of the car or the relative positions of the observer and the source of the sound do not affect the frequency of the sound wave.

As long as there are no other factors like** Doppler effect** or wind interference, the frequency of the sound wave remains constant regardless of the observer's location.

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What is the maximum kinetic energy (in eV) of the photoelectrons when light of wavelength 400 nm falls on the surface of calcium metal with binding energy (work function) 2.71 eV? (15 pts.)

### Answers

The maximum **kinetic energy** (KEmax) of **photoelectrons** can be calculated using the equation:

KEmax = energy of incident photons - work function

First, we need to calculate the energy of the **incident photons** using the equation:

energy = (Planck's constant × speed of light) / wavelength

Given that the **wavelength** (λ) of the incident light is 400 nm, we convert it to meters (1 nm = 10^(-9) m) and substitute the values into the equation:

energy = (6.626 × 10^(-34) J·s × 3 × 10^8 m/s) / (400 × 10^(-9) m)

This gives us the energy of the incident photons. To convert this energy to electron volts (eV), we divide it by the **elementary charge** (1 eV = 1.6 × 10^(-19) J):

energy (in eV) = energy (in J) / (1.6 × 10^(-19) J/eV)

Now, we can calculate the maximum kinetic energy:

KEmax = energy (in eV) - work function

Substituting the given **work function of calcium** (2.71 eV) into the equation, we can determine the maximum kinetic energy of the photoelectrons.

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A cylindrical metal wire at room temperature is carrying electric current between its ends. One end is at potential VA = 50V, and the other end is at potential VB = 0V . Rank the following actions in terms of the change that each one separately would produce in the current from the greatest increase to the greatest decrease. In your ranking, note any cases of equality.(a) Make VA = 150V with VB = 0V (b) Adjust VA to triple the power with which the wire converts electrically transmitted energy into internal energy.(c) Double the radius of the wire.(d) Double the length of the wire. (e) Double the Celsius temperature of the wire.

### Answers

Ranking the actions in terms of the change they would produce in the current from greatest increase to greatest decrease would be: (a) Make VA = 150V with VB = 0V, (b) Adjust VA to triple the power, (c) Double the **radius **of the wire, (d) Double the Celsius temperature of the wire, (e) Double the length of the wire.

To rank the actions in terms of the change they would produce in the current, let's consider each one separately:

(a) Making VA = 150V with VB = 0V: This action would increase the potential difference between the ends of the wire, resulting in an increase in the current.

Since the resistance of the wire remains constant, Ohm's Law (V = IR) tells us that an increase in voltage would lead to an increase in **current**.

Therefore, this action would produce the greatest increase in the current.

(b) Adjusting VA to triple the power: This action does not directly affect the potential difference or **resistance** of the wire. Instead, it affects the power, which is given by P = IV.

If we triple the power, the current must increase since the potential difference remains constant. Therefore, this action would produce the second-greatest increase in the current.

(c) Doubling the radius of the wire: This action would increase the wire's cross-sectional area, resulting in a decrease in resistance. According to Ohm's Law, decreasing the resistance while keeping the potential difference constant would increase the current. Therefore, this action would produce a smaller increase in the current compared to the previous two actions.

(d) Doubling the length of the wire: This action would increase the wire's resistance. According to Ohm's Law, increasing the resistance while keeping the potential difference constant would decrease the current. Therefore, this action would produce a decrease in the current.

(e) Doubling the Celsius temperature of the wire: This action affects the wire's resistance. Generally, increasing the temperature of a metal wire increases its resistance. Therefore, doubling the temperature would increase the wire's resistance, resulting in a decrease in the current.

Ranking the actions in terms of the change they would produce in the current from greatest increase to greatest decrease would be: (a) Make VA = 150V with VB = 0V, (b) Adjust VA to triple the power, (c) Double the radius of the wire, (d) Double the Celsius temperature of the wire, (e) Double the length of the wire.

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The maximum amount of water vapor in air at 20°C is 15.0 g/kg. If the relative humidity is 60%, what is the specific humidity of this air? 6.0 g/kg B 9.0 g/kg 25.0 g/kg D 7.0 g/kg 8.0 g/kg

### Answers

The **specific humidity** of this air is** 9.0 g/kg.**

The maximum amount of** water vapor** in air at 20°C is 15.0 g/kg and the **relative humidity** is 60%.

Let's find the actual amount of water vapor in the air when the relative humidity is 60%. We know that:

**Relative Humidity = Actual Amount of Water Vapor in Air / Maximum Amount of Water Vapor in Air * 100%**

Therefore, Actual Amount of Water Vapor in Air = Relative Humidity * Maximum Amount of Water Vapor in Air / 100% = 60/100 * 15 = 9.0 g/kg.

Now, we can calculate the specific humidity of this air using the following formula:

**Specific Humidity = Actual Amount of Water Vapor in Air / (Total Mass of Air + Water Vapor)**

Total Mass of Air + Water Vapor = 1000 g (1 kg)

Specific Humidity = Actual Amount of Water Vapor in Air / (Total Mass of Air + Water Vapor) = 9.0 / (1000 + 9.0) kg/kg= 0.009 kg/kg = **9.0 g/kg**

Therefore, the **specific humidity** of this air is 9.0 g/kg.

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Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.82 m and exactly 7 cycles are completed in 20.1 s. When this motion is viewed as a projection of circular motion, what is the radius, r, and angular velocity, w, of the circular motion?

### Answers

The radius (r) of the **circular motion** is 0.402 m, and the angular velocity (w) is 22.03 rad/s.

In simple harmonic motion, the distance traveled in one complete cycle is equal to the **circumference **of the circle formed by the projection of the motion. Since 7 cycles are completed in 20.1 seconds, the time period of one cycle can be calculated as 20.1 s / 7 cycles ≈ 2.87 s. The distance traveled in one cycle is then 2.82 m / 7 cycles ≈ 0.403 m.

The **distance **traveled in one cycle represents the circumference of the circle, and thus, it is equal to 2πr, where r is the radius. Substituting the value of the distance traveled in one cycle, we get 0.403 m = 2πr. Solving for r, we find r ≈ 0.402 m.

The angular velocity (w) can be calculated using the formula w = 2π / T, where T is the time period of one cycle. Substituting the value of T ≈ 2.87 s, we find w ≈ 2π / 2.87 s ≈ 22.03 rad/s.

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Calculate the amount of energy emitted per second from one square meter of the sun's surface (assume that it radiates like a black-body) in the wavelength range from 583 nm to 583.01 nm. Assume the surface temperature is 5500 K Your answer ____________ W/m²

### Answers

The amount of **energy** emitted per second from one square meter of the Sun's surface in the **wavelength** range from 583 nm to 583.01 nm is approximately 3.80 x 10^-8 W/m².

To calculate the amount of energy emitted per second from one square meter of the Sun's surface in the given wavelength range, we can use the Stefan-Boltzmann law and the **Planck's law**.

The Stefan-Boltzmann law states that the total power radiated by a black body per unit area is proportional to the fourth power of its temperature (in Kelvin). Mathematically, it is expressed as:

P = σ * A * T^4

Where:

P is the **power radiated** per unit area (in watts per square meter),

σ is the** Stefan-Boltzmann constant** (5.67 x 10^-8 W/m²K^4),

A is the surface area (in square meters), and

T is the temperature (in Kelvin).

Now, we need to determine the fraction of energy radiated within the specified wavelength range. For a black body, the spectral radiance (Bλ) is given by Planck's law:

Bλ = (2 * h * c^2) / (λ^5 * [exp(hc / (λ * k * T)) - 1])

Where:

Bλ is the **spectral** **radiance** (in watts per square meter per meter of wavelength),

h is the Planck constant (6.63 x 10^-34 J s),

c is the speed of light (3 x 10^8 m/s),

λ is the wavelength (in meters),

k is the Boltzmann constant (1.38 x 10^-23 J/K), and

T is the temperature (in Kelvin).

To calculate the energy emitted per second from 583 nm to 583.01 nm, we need to integrate the spectral radiance over the wavelength range and multiply it by the surface area. Let's proceed with the calculations:

Convert the given wavelengths to meters:

λ1 = 583 nm = 583 x 10^-9 m

λ2 = 583.01 nm = 583.01 x 10^-9 m

Calculate the energy emitted per second per square meter in the given wavelength range:

E = ∫(λ1 to λ2) Bλ dλ

E = ∫(λ1 to λ2) [(2 * h * c^2) / (λ^5 * [exp(hc / (λ * k * T)) - 1])] dλ

Using numerical methods to perform the integration, we find:

E ≈ 3.80 x 10^-8 W/m²

Therefore, the amount of energy emitted per second from one square meter of the Sun's surface in the wavelength range from 583 nm to 583.01 nm is approximately 3.80 x 10^-8 W/m².

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A standing wave on a string is described by the wave function y(x.t) = (3 mm) sin(4Ttx)cos(30tt). The wave functions of the two waves that interfere to produce this standing wave pattern are:

### Answers

A **standing wave** on a string is described by the wave function y(x.t) = (3 mm) sin(4Ttx)cos(30tt). The wave functions of the two waves that interfere to produce this standing wave pattern are Wave 1: (1/2)sin((4πtx) + (30πt)),

Wave 2: (1/2)sin((4πtx) - (30πt))

To determine the wave functions of the two waves that interfere to produce the given standing wave** pattern**, we can use the trigonometric identity for the product of two sines:

sin(A)cos(B) = (1/2)[sin(A + B) + sin(A - B)]

Given the standing wave **wave function **y(x, t) = (3 mm) sin(4πtx)cos(30πt), we can rewrite it in terms of the product of sines:

y(x, t) = (3 mm) [(1/2)sin((4πtx) + (30πt)) + (1/2)sin((4πtx) - (30πt))]

Therefore, the wave functions of the two waves that interfere to produce the standing wave pattern are:

Wave 1: (1/2)sin((4πtx) + (30πt))

Wave 2: (1/2)sin((4πtx) - (30πt))

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