Physics High School

## Answers

**Answer 1**

The magnitude of the** friction force** acting on the car is 3434 N when a 500 kg car is parked on a hill with a 5° slope.

The **magnitude** of the friction force acting on the car can be calculated using the formula Ff = μFn, where Ff is the friction force, μ is the coefficient of friction, and Fn is the normal force. In this case, the car is parked on a hill with a 5° slope, so the **normal force** acting on the car is equal to its weight, which can be calculated as Fg = mg = 500 kg x 9.81 m/s² = 4905 N.

The **coefficient **of friction depends on the surfaces in contact. Let's assume the car's tires are made of rubber and the road surface is concrete. In this case, the coefficient of static friction between rubber and concrete is typically between 0.7 and 0.9. Let's take a conservative estimate of μ = 0.7.

Now we can calculate the **friction force** as Ff = μFn = 0.7 x 4905 N = 3434 N. Therefore, the magnitude of the friction force acting on the car is approximately 3434 N, which is the force that opposes the car's motion down the hill due to gravity.

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## Related Questions

A rotating flywheel of a diameter 40.0 cm uniformly acceleratesfrom rest to 250 rad/s in 15.0 s. (a) Find its angularacceleration. (b) Find the linear velocity of a pointon the rim of the wheel after 15.0 s. (c) How manyrevolutions does the wheel make during the 15.0 s?

### Answers

(a)the angular **acceleration** of the flywheel is 16.7 rad/s^2.

(b)the linear velocity of a point on the rim of the wheel after 15.0 s is 50 m/s.

(c)the wheel makes **approximately** 29.8 revolutions during the 15.0 s.

(a) The initial angular velocity of the flywheel, ω0 = 0. The final angular velocity, ω = 250 rad/s. The time, t = 15.0 s. Using the formula,

ω = ω0 + αt

where α is the **angular** acceleration, we can solve for α:

α = (ω - ω0)/t = 250 rad/s / 15.0 s = 16.7 rad/s^2

Therefore, the angular acceleration of the flywheel is 16.7 rad/s^2.

(b) The **linear** velocity, v, of a point on the rim of the wheel is given by:

v = rω. where r is the radius of the wheel. Substituting r = 0.2 m and ω = 250 rad/s, we get:

v = (0.2 m)(250 rad/s) = 50 m/s

Therefore, the linear velocity of a point on the rim of the wheel after 15.0 s is 50 m/s.

(c) The number of revolutions made by the wheel during the 15.0 s can be **calculated** using the formula: θ = ω0t + (1/2)αt^2

where θ is the angular displacement of the wheel. Since the wheel starts from rest, ω0 = 0. Also, the final angular velocity, ω, is given by:

ω^2 = ω0^2 + 2αθ

Solving for θ, we get: [tex]θ = (ω^2 - ω0^2) / 2α = (250^2 - 0^2) / (2 x 16.7) = 187.1 rad[/tex]

The number of revolutions, N, made by the wheel can be calculated as:

N = θ / 2π = 187.1 rad / (2π) = 29.8 revolutions (approx)

**Therefore**, the wheel makes approximately 29.8 revolutions during the 15.0 s.

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a. Calculate the centripetal force exerted on a vehicle of mass m=1630 kg that is moving at a speed of 13.9 m/s around a curve of radius R=385 m.

b. Which force plays the role of the centripetal force in this case?

a. force of static friction

b. spring force

c. gravitational force

d. normal force

e. tension force

### Answers

a. Fc = (m * v^2) / R

where m is the mass of the vehicle, v is the velocity of the vehicle, and R is the radius of the curve.

Plugging in the values, we get:

Fc = (1630 kg * (13.9 m/s)^2) / 385 m

Fc = 8206.73 N

Therefore, the centripetal force exerted on the vehicle is 8206.73 N.

b. The force that plays the role of the centripetal force in this case is the force of static friction.

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For a force F of constant magnitude, the torque it applies on an object must increases if O the axis of rotation of the object approaches the point of application of F. O the moment arm of F increases. O the angle of application of F increases. O the line of action of F increases.

### Answers

**Torque** increases when the axis of rotation approaches the point where force is applied, the moment arm increases, or the angle of application increases, while the line of **action of the force** does not directly affect the torque but can impact the moment arm and angle of application.

The torque increases if:

1. The axis of rotation of the object approaches the point of application of F.

2. The **moment arm** of F increases.

3. The angle of application of F increases.

The torque (τ) can be calculated using the formula τ = F × r × sinθ, where F is the force, r is the moment arm (the distance between the **axis of rotation** and the point of application of the force), and θ is the angle between the force and the moment arm.

From this formula, it is evident that the torque increases when the moment arm (r) increases or when the angle (θ) increases. Additionally, as the axis of rotation approaches the point of application of the force, the moment arm will increase, resulting in an increase in torque.

In summary, the **torque** applied by a force F of constant magnitude on an object increases when the axis of rotation approaches the point of application of F, the moment arm of F increases, or the angle of application of F increases. The line of action of F does not affect the torque directly but may influence the moment arm and the angle of application.

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6 silicon electron mobility at t=300k is u=970 a) calculate diffusion coefficient electrons

### Answers

The diffusion **coefficient **of electrons is 4.28 x 10⁻¹⁰ [tex]m^2/s.[/tex]

**What will be diffusion coefficient electrons?**

To calculate the diffusion coefficient of **electrons**, we can use the Einstein relation:

D = kT/u

where D is the diffusion coefficient, k is Boltzmann's constant (1.38 x 10⁻²³ J/K), T is the **temperature **in Kelvin, and u is the electron mobility.

Plugging in the values given in the problem:

u = 970 [tex]cm^2/Vs[/tex] (note that the units need to be converted to [tex]cm^2/Vs[/tex])

T = 300 K

k = 1.38 x 10⁻²³ J/K

Converting the units of electron **mobility **from [tex]cm^2/Vs[/tex] to [tex]m^2/s[/tex] gives:

u = 9.7 x 10⁻³ [tex]m^2/s[/tex]

Now we can calculate the diffusion coefficient:

D = kT/u

D = (1.38 x 10⁻²³ J/K) x (300 K) / (9.7 x 10⁻³ [tex]m^2/s[/tex])

D = 4.28 x 10⁻¹⁰ [tex]m^2/s[/tex]

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Calculate the power per square meter (in kW/m2) reaching Earth's upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00 ✕ 1026W.)

### Answers

The **power from the sun** reaching Earth's upper atmosphere is 1.42 x 10³ kW/m².

To calculate the power per square meter reaching Earth's upper atmosphere from the Sun, we need to use the **inverse square law**.

The power output of the Sun is given as 4.00 x 10²⁶ W.

The distance between the Sun and the Earth varies throughout the year, but on average, it is about 149.6 million kilometers (9.3 x 10⁷ miles).

Using the formula for the **surface area of a sphere**, we can find the total surface area of the imaginary sphere with a radius equal to the distance between the Sun and the Earth.

The surface area of a sphere = 4πr²

The surface area of the sphere with a radius of 149.6 million km:

A = 4 x 3.1416 x (149.6 x 10⁹)²

A = 2.827 x 10²³ m²

Now, we can calculate the power per square meter reaching Earth's upper atmosphere by dividing the total power output of the Sun by the total surface area of the sphere.

Power per square meter = Power output of the Sun / Total surface area of the sphere

= (4.00 x 10²⁶ W) / (2.827 x 10²³ m²)

= 1.42 x 10³ kW/m²

Therefore, the power per square meter reaching **Earth's upper atmosphere** from the Sun is 1.42 x 10³ kW/m².

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How much impulse (in magnitude) stops an object with mass m= 10 kg sliding at v = 3m/s? a. 30 Ns

b. 0.6 kg m/s

c. 60 kg m/s

d. 10 Ns

### Answers

a.) The impulse required to stop an object with a **mass **of 10 kg and** velocity **of 3 m/s is 30 Ns.

Impulse is defined as the change in **momentum **of an object, which is equal to the force applied multiplied by the time it acts. In this case, the object has a mass of 10 kg and is** moving** with a velocity of 3 m/s. To stop the object, a force needs to be applied in the opposite direction of its motion. The impulse required to stop the object can be calculated by multiplying the **force **applied with the time it acts. Since the impulse is equal to the change in momentum, it can be calculated using the equation [tex]J = ∆p = mv_f - mv_i.[/tex] Solving for the force, we get[tex]F = J/t = ∆p/t = m(v_f - v_i)/t = m*a,[/tex] where a is the acceleration produced by the force. Therefore, the impulse required to stop the object is 30 Ns, which is **equivalent** to the force of 30 N acting for 1 second.

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a hair breaks under a tension of 1.2 n. what is the diameter of the hair? the tensile strength is 2.2 ✕ 108 pa.

### Answers

The diameter of the hair is approximately 3.12 micrometers.

To find the diameter of the hair, we can use the formula for **tensile strength**:

Tensile strength = Force / Area

We know that the** tension force** is 1.2 N and the tensile strength is 2.2 ✕ 108 Pa. We can rearrange the formula to solve for the area (which will give us the cross-sectional area of the hair):

**Area** = Force / Tensile strength

Substituting the values we have:

Area = 1.2 N / 2.2 ✕ 108 Pa

Area = 5.45 ✕ 10^-9 m^2

Now, we can use the formula for the area of a circle to find the diameter:

Area = π/4 ✕ diameter^2

Solving for diameter:

**diameter **= √(4 ✕ Area / π)

Substituting the value we found for the area:

diameter = √(4 ✕ 5.45 ✕ 10^-9 / π)

diameter = 3.12 ✕ 10^-6 m

Therefore, the diameter of the hair is approximately 3.12 micrometers.

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If the S/N of the input signal is 4 and the intelligence signal is 10 kHz, what is the deviation? a. 100 kHz b. 145 kHz c. 160 kHz d. 200 kHz

### Answers

If the S/N of the input signal is 4 and the intelligence signal is 10 kHz, what is the **deviation** is 200 kHz. The correct option d.

**Deviation** can be calculated using the formula:

Deviation = (S/N) x (Intelligence signal frequency)

Substituting the given values:

Deviation = (4) x (10 kHz) = 40 kHz

However, this only gives us the **peak deviation**. In frequency modulation, the actual deviation is determined by the modulation index, which is dependent on the amplitude of the intelligence signal.

Assuming a maximum modulation index of 5 (which is a typical value for FM broadcasting), the actual deviation can be calculated as: **Actual deviation = (Modulation index) x (Peak deviation) **

Actual deviation = (5) x (40 kHz) = 200 kHz

The correct option d.

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How many photons per second enter one eye if you look directly at a 100 W light bulb 2.00 m away? Assume a pupil diameter of 4.00 mm and a wavelength of 600 nm. How many photons per second enter your eye if a 1.00 m W laser beam is directed into your eye? λ=633nm)

### Answers

The number of **photons** per second that enter the eye can be calculated using the formula:

N = (P / A) x (t / h) x (1 / E)

where:

P = power of the light source (in watts)

A = area of the pupil (in square meters)

t = transmission coefficient of the cornea and lens (assumed to be 0.95)

h =** Planck's constant **(6.626 x 10[tex]^-34[/tex] joule-seconds)

E = energy per photon (in joules)

For the 100 W light bulb:

P = 100 W

A = π (0.002 m)^2 = 1.2566 x 10[tex]^-5 m^2[/tex] (assuming the pupil is circular)

t = 0.95 (given)

h = 6.626 x 10[tex]^-34[/tex] J·s (given)

λ = 600 nm = 6.00 x 10[tex]^-7 m[/tex] (given)

c = speed of light = 3.00 x 10m/s (assumed)

E = hc / λ = (6.626 x 10[tex]^-34[/tex] J·s) x (3.0[tex]^8[/tex]0 x 10[tex]^8[/tex] m/s) / (6.00 x 10[tex]^-7 m[/tex]) = 3.31 x 10[tex]^-19[/tex] J

**Plugging in** the values:

N = (100 W / 1.2566 x 10[tex]^-5 m^2[/tex]) x (0.95) x (1 s / 6.626 x 10[tex]^-34[/tex] J·s) x (1 / 3.31 x 10[tex]^-19[/tex] J)

= 7.70 x 10^16 photons/s

Therefore, about 7.70 x 10[tex]^16[/tex] photons per second enter one eye when looking directly at a 100 W light bulb from a distance of 2.00 m.

For the 1.00 mW laser beam:

P = 1.00 x 10[tex]^-3[/tex] W

A = π (0.002 m[tex])^2[/tex] = 1.2566 x 10[tex]^-5 m^2[/tex] (assuming the pupil is circular)

t = 0.95 (given)

h = 6.626 x 10[tex]^-34[/tex]J·s (given)

λ = 633 nm = 6.33 x 10[tex]^-7[/tex] m (given)

**c = speed of light **= 3.00 x 10[tex]^8[/tex] m/s (assumed)

E = hc / λ = (6.626 x 10[tex]^-34[/tex] J·s) x (3.00 x 10[tex]^8[/tex]m/s) / (6.33 x 10[tex]^-7[/tex]m) = 3.14 x 10[tex]^-19[/tex] J

Plugging in the values:

N = (1.00 x 10[tex]^-3[/tex]W / 1.2566 x 10[tex]^-5 m^2[/tex]) x (0.95) x (1 s / 6.626 x 10[tex]^-34[/tex] J·s) x (1 / 3.14 x 10[tex]^-19[/tex]J)

= 7.17 x 10^[tex]12[/tex] photons/s

Therefore, about 7.17 x 10[tex]^12[/tex] photons per second enter your eye if a 1.00 mW l**aser beam** with a wavelength of 633 nm is directed into your eye.

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The number of **photons** per second that enter the eye can be calculated using the formula:

N = (P / A) x (t / h) x (1 / E)

where:

P = power of the light source (in watts)

A = area of the pupil (in square meters)

t = transmission coefficient of the cornea and lens (assumed to be 0.95)

h =** Planck's constant **(6.626 x 10[tex]^-34[/tex] joule-seconds)

E = energy per photon (in joules)

For the 100 W light bulb:

P = 100 W

A = π (0.002 m)^2 = 1.2566 x 10[tex]^-5 m^2[/tex] (assuming the pupil is circular)

t = 0.95 (given)

h = 6.626 x 10[tex]^-34[/tex] J·s (given)

λ = 600 nm = 6.00 x 10[tex]^-7 m[/tex] (given)

c = speed of light = 3.00 x 10m/s (assumed)

E = hc / λ = (6.626 x 10[tex]^-34[/tex] J·s) x (3.0[tex]^8[/tex]0 x 10[tex]^8[/tex] m/s) / (6.00 x 10[tex]^-7 m[/tex]) = 3.31 x 10[tex]^-19[/tex] J

**Plugging in** the values:

N = (100 W / 1.2566 x 10[tex]^-5 m^2[/tex]) x (0.95) x (1 s / 6.626 x 10[tex]^-34[/tex] J·s) x (1 / 3.31 x 10[tex]^-19[/tex] J)

= 7.70 x 10^16 photons/s

Therefore, about 7.70 x 10[tex]^16[/tex] photons per second enter one eye when looking directly at a 100 W light bulb from a distance of 2.00 m.

For the 1.00 mW laser beam:

P = 1.00 x 10[tex]^-3[/tex] W

A = π (0.002 m[tex])^2[/tex] = 1.2566 x 10[tex]^-5 m^2[/tex] (assuming the pupil is circular)

t = 0.95 (given)

h = 6.626 x 10[tex]^-34[/tex]J·s (given)

λ = 633 nm = 6.33 x 10[tex]^-7[/tex] m (given)

**c = speed of light **= 3.00 x 10[tex]^8[/tex] m/s (assumed)

E = hc / λ = (6.626 x 10[tex]^-34[/tex] J·s) x (3.00 x 10[tex]^8[/tex]m/s) / (6.33 x 10[tex]^-7[/tex]m) = 3.14 x 10[tex]^-19[/tex] J

Plugging in the values:

N = (1.00 x 10[tex]^-3[/tex]W / 1.2566 x 10[tex]^-5 m^2[/tex]) x (0.95) x (1 s / 6.626 x 10[tex]^-34[/tex] J·s) x (1 / 3.14 x 10[tex]^-19[/tex]J)

= 7.17 x 10^[tex]12[/tex] photons/s

Therefore, about 7.17 x 10[tex]^12[/tex] photons per second enter your eye if a 1.00 mW l**aser beam** with a wavelength of 633 nm is directed into your eye.

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a capacitor of 10.0 f and a resistor of 120 ω are quickly connected in series to a battery of 6.00 v. what is the charge on the capacitor 0.00100 s after the connection is made?

### Answers

The charge on the** capacitor** 0.00100 s after the connection is made is approximately 49.3 μC.

To calculate the charge on the capacitor at a specific time, you can use the formula:

Q(t) = C * V * (1 - e^(-t / (R * C)))

where Q(t) is the charge at time t, C is the capacitance (10.0 F), V is the **battery voltage **(6.00 V), R is the resistance (120 Ω), and t is the time (0.00100 s).

1. Calculate the product of **resistance** and capacitance: RC = 120 Ω * 10.0 F = 1200 s.

2. Calculate the negative exponent term: -t / (RC) = -0.00100 s / 1200 s = -0.000000833.

3. Evaluate e^(-t / (RC)): e^(-0.000000833) ≈ 0.99917.

4. Subtract the result from step 3 from 1: 1 - 0.99917 ≈ 0.00083.

5. Multiply the result from step 4 by the product of** capacitance** and voltage: 0.00083 * (10.0 F * 6.00 V) ≈ 49.3 μC.

So, the charge on the capacitor 0.00100 s after the connection is made is approximately 49.3 μC.

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To A The vector A has a magnitude of 15.0, the vector B has a magnitude of 12.0, and the angle between them is € = 25.00 Determine AX B. 76.1 out of page 180 into page 76.1 into page 180 out of page 163 out of page 163 into page

### Answers

So, AxB has a **magnitude **of 76.1 and its **direction **is "out of the page."

Since we need to **determine **the cross product of vectors A and B (written as AxB), we'll use the given information about their **magnitudes **and the angle between them.

Step 1: Note the given information

- Magnitude of vector A: 15.0

- Magnitude of vector B: 12.0

- Angle between A and B: θ = 25.0°

Step 2: Use the formula for cross product magnitude

The magnitude of AxB can be found using the **formula**: |AxB| = |A| * |B| * sin(θ), where |A| and |B| are the magnitudes of vectors A and B, and θ is the angle between them.

Step 3: Calculate the magnitude of AxB

- |AxB| = (15.0) * (12.0) * sin(25.0°)

- |AxB| ≈ 76.1

Step 4: Determine the direction of AxB

Since the cross product is a vector that is perpendicular to both vectors A and B, it will be either "out of the page" or "into the page." To determine this, use the right-hand rule: point your thumb in the direction of A, your index finger in the direction of B, and your middle **finger **will point in the direction of AxB. In this case, the direction is "out of the page."

So, AxB has a magnitude of 76.1 and its direction is "out of the page."

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a particular metallic element, m, has a heat capacity of 0.36 j • g-1 • k-1. it forms an oxide that contains 2.90 g of m per gram of oxygen.

### Answers

The** molar mass **of the metallic element m is 2.90 g/mol.

**What is Element?**

An element is a pure substance that consists of only one type of atom. It is the simplest form of matter that cannot be broken down into simpler substances by chemical means. Each element is uniquely characterized by its **atomic number**, which represents the number of** protons** in the **nucleus **of its atoms, and its **atomic** symbol, which is a shorthand notation for the element.

We can use the heat capacity formula to calculate the molar mass of m:

C = (moles of substance) * (molar **heat capacity)**

Rearranging the formula to solve for moles of substance:

moles of substance = C / molar heat capacity

Now we can substitute the given values and solve for the moles of m in the oxide:

This means that 2.90 g of m is equivalent to 1 mole of m.

To find the **molar mass **of m, we divide the mass of m by the moles of m:

molar mass of m = mass of m / moles of m

molar mass of m = 2.90 g / 1 mole = 2.90 g/mol

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Flip a biased coin 100 times. On each flip, P[H] =p. LetXi denote the number of heads that occur on flip i.

a.) What is PX33 (x)?

b.) Are X1 and X2 independent? why?

Define Y = X1 + X2 + ....... +X1000

c.) What is PY (y)

d.) E[Y] and Var [Y].

### Answers

These formulas provide a way to calculate the mean and **variability **of Y, which are important measures for understanding the behavior of **random variables**.

a) The probability mass function for [tex]X_{33}[/tex] is given by [tex]PX_{33}[/tex](x) = (100 choose x) * [tex]p^x[/tex] *[tex](1-p)^(100-x).[/tex]

b) No, X1 and X2 are not **independent** because the outcome of the first coin flip can affect the outcome of the second coin flip.

c) The distribution of Y is a binomial **distribution** with parameters n = 1000 and p. Therefore, the probability mass function for Y is given by PY(y) = (1000 choose y) * [tex]p^y[/tex] * [tex](1-p)^(1000-y).[/tex]

d) The **expected value **of Y is E[Y] = np = 1000p, and the variance of Y is Var[Y] = np(1-p) = 1000p(1-p).

Variability refers to the degree to which data or **observations** fluctuate or vary from the mean or expected value. It is a statistical concept that measures the dispersion of data points in a dataset. A dataset with high variability indicates that the data points are widely spread out, while a low variability indicates that the data points are **clustered** around the mean.

There are different measures of variability, **including** **range**, variance, and standard deviation. Range is the difference between the maximum and minimum values in a dataset, while variance and standard deviation measure the average squared deviation from the mean. These measures are important in many fields, such as finance, economics, and psychology, as they help to understand the spread of data and assess the reliability of **statistical** estimates.

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The gap in the outer and inner walls of a calorimeter (the device to measure heat) is filled with air because ita. allows only the heat to flow from inside to outside of the calorimeterb. restricts only the heat to flow from inside to outside of the calorimeterc. restricts the flow of heat from inside to outside and outside to inside of the calorimeter

### Answers

The gap in the inner walls as well as outer walls of the **calorimeter** is filled with air because it restricts the flow of the heat from inside to outside as well as outside to inside of the calorimeter. Option C is correct.

The gap in the outer and** inner walls** of a calorimeter is typically filled with air to act as an insulator. Air is a poor conductor of heat, meaning that it restricts the flow of heat from inside to outside and outside to inside of the calorimeter. This helps to maintain the** temperature** inside the calorimeter and prevents heat from escaping or entering the calorimeter too quickly.

Hence, C. is the correct option.

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A thin semicircular rod has a total charge +Q uniformly distributed along it. A negative point charge - Q is placed as shown. A test charge +q is placed at point C (point C is equidistant from all points on the rod.). Let F_P and F_R represent the force on the test . charge and the rod respectively. Is the magnitude of the net force on +q than, less than, or equal to the magnitude of Explain. A second negative point charge -Q is placed as shown. Is the magnitude of the net electric force on +q greater than, less than, or equal to the magnitude of the net electric force on +q in part b? Explain.

### Answers

The magnitude of the net force on +q will be less.

The magnitude of the **net electric force** on +q will be greater than the magnitude of the net electric force on +q.

In the given scenario, the force F_P on the test charge +q will be **attractive** towards the negative point charge -Q. The force F_R on the rod will also be attractive towards -Q due to the presence of the negative charge.

However, the force on the** test charge** +q will be less than the force on the rod as the test charge is equidistant from all points on the rod and experiences equal but opposite forces from opposite points on the rod, resulting in cancellation.

When a second negative point charge -Q is placed, the net force on the test charge +q will be greater than the net force in part b. This is because the presence of the second negative charge will cause a repulsive force on the first negative charge, which will in turn reduce the attractive force on the test charge +q towards the negative charge -Q.

As a result, the **net electric force **on the test charge +q will be greater due to the reduced attractive force towards the negative charge -Q.

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Directions: Match the terms in the second column with their related terms in the first column by writing the

correct letter in the space provided.

1. electromagnetic wave

2. radar

3. c

4. photon

5. infrared waves

6. frequency

7. ultraviolet waves

8. microwaves

9. CRT

10. pager

11. moving electric field

12. carrier wave

a. wavelength

b. television

c. behaves as a particle

d. oscillating electric charge

e. moving magnetic field

f. sunburn

g. frequency modulation

h. tracking storms

i. cooking food

j. speed of light

k. TV remote control

1. unique identification

number

### Answers

**Electromagnetic **Wave - oscillating electric charge

Radar - tracking storms

c - Velocity of light

Photon - behaves as a particle

infrared waves - TV remote control

Frequency - Wavelength

ultraviolet waves - sun burn

**Microwaves **- Cooking Food

CRT - Television

pager - unique identification number

moving electric field - moving magnetic field

carrier wave - frequency modulation

In physics, **electromagnetic **radiation (EMR) is made up of electromagnetic (EM) field waves that travel across space carrying velocity and electromagnetic radiant energy. Radio waves, **microwaves**, infrared, (visible) light, ultraviolet, X-rays, and gamma rays are all examples of electromagnetic radiation.

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A Hobbit Home is surely a cooler design than your home, but why should Bilbo consider rethinking the design of the front door? A larger handle is needed when pulling at the center of mass Less leverage when pulling would be better That circular door has more rotational inertia than a rectangular door would The handle should be lower for smaller hobbits to pull and creating less rotational inertia More leverage when pulling would be better

### Answers

While a Hobbit home may be a cooler design than a regular home, there are still functional considerations that should be taken into account when designing the front door. A larger handle, more **leverage**, and a lower handle would all make it easier to open the door and reduce the amount of force required to get it moving.

Bilbo may want to consider rethinking the design of the front door for a few reasons:

A larger** handle** is needed when pulling at the center of mass: The circular shape of the door means that the center of mass is not in the middle of the door. This makes it harder to pull the door open, especially if the handle is too small. A larger handle would provide more surface area to grip and make it easier to open the door.More leverage when pulling would be better: The **circular shape **of the door also means that there is less leverage when pulling the door open. A rectangular door with a longer handle would provide more leverage, making it easier to open the door.The handle should be lower for smaller hobbits to pull and creating less **rotational inertia**: The circular shape of the door creates more rotational inertia than a rectangular door would. This means that it takes more force to get the door moving, and once it is moving, it is harder to stop. A lower handle would make it easier for smaller hobbits to pull the door open and would also create less rotational inertia.

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Suppose the tilt of Earth's equator relative to its orbit were 30 ° instead of 23.5°. At what latitudes would the Arctic and Antarctic Circles be located?

_____ degrees latitude

### Answers

The new position of the** Arctic** and **Antarctic** Circles would be located at **30° North and 30° South**, respectively.

If the tilt of **Earth's equator** relative to its orbit were 30° instead of 23.5°, the location of the Arctic and Antarctic Circles would be affected. The Arctic Circle is defined as the **latitude** above which the sun does not set on the summer solstice and does not rise on the winter solstice. The same applies to the Antarctic Circle but in the Southern Hemisphere.

Currently, the Arctic Circle is located at 66.5° North and the Antarctic Circle is located at 66.5° South. If the** tilt** of Earth's equator relative to its** orbit** were 30°, the position of the Arctic and Antarctic Circles would be closer to the equator.

To determine the new **position **of the Arctic and Antarctic Circles, we can use the following formula:

θ = 90° - ϕ

where θ is the angle between the **axis of rotation** and the line perpendicular to the plane of the ecliptic, and ϕ is the tilt of the Earth's axis relative to the plane of the ecliptic.

If the tilt were 30°, then θ would be:

θ = 90° - 30° = 60°

Therefore, the new **latitude** of the Arctic and Antarctic Circles would be:

ϕ = 90° - 60° = 30°

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The new position of the** Arctic** and **Antarctic** Circles would be located at **30° North and 30° South**, respectively.

If the tilt of **Earth's equator** relative to its orbit were 30° instead of 23.5°, the location of the Arctic and Antarctic Circles would be affected. The Arctic Circle is defined as the **latitude** above which the sun does not set on the summer solstice and does not rise on the winter solstice. The same applies to the Antarctic Circle but in the Southern Hemisphere.

Currently, the Arctic Circle is located at 66.5° North and the Antarctic Circle is located at 66.5° South. If the** tilt** of Earth's equator relative to its** orbit** were 30°, the position of the Arctic and Antarctic Circles would be closer to the equator.

To determine the new **position **of the Arctic and Antarctic Circles, we can use the following formula:

θ = 90° - ϕ

where θ is the angle between the **axis of rotation** and the line perpendicular to the plane of the ecliptic, and ϕ is the tilt of the Earth's axis relative to the plane of the ecliptic.

If the tilt were 30°, then θ would be:

θ = 90° - 30° = 60°

Therefore, the new **latitude** of the Arctic and Antarctic Circles would be:

ϕ = 90° - 60° = 30°

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A steel cable initially 7m long is used to strap three logs of timber onto a vehicle. The cable is under a tension of 400N. If the diameter of the cable is 4cm, i) By how much has it to be extended under this tension. ii) How much work has been done in producing this extension. (Young's modulus for steel is 2.0×10^11 N\m^2)

### Answers

**Answer:**

i) The extension of the cable can be calculated using the formula:

ΔL/L = F/(A*Y)

where ΔL is the extension in length, L is the original length, F is the tension force, A is the cross-sectional area of the cable, and Y is the Young's modulus of steel.

First, we need to calculate the cross-sectional area of the cable:

A = πr^2 = π(0.02m)^2 = 0.00126 m^2

Now we can calculate the extension:

ΔL/L = 400N/(0.00126m^2 * 2.0×10^11 N/m^2) = 1.5873×10^-6

ΔL = (1.5873×10^-6)(7m) = 1.11×10^-5 m

So the cable has to be extended by 1.11×10^-5 m under this tension.

ii) The work done in producing this extension can be calculated using the formula:

W = (1/2)FΔL

where W is the work done, F is the tension force, and ΔL is the extension in length.

Substituting the given values:

W = (1/2)(400N)(1.11×10^-5 m) = 2.22×10^-3 J

Therefore, the work done in producing this extension is 2.22×10^-3 J.

**Explanation:**

A resistor with a 15.0-V potential difference across its ends develops thermal energy at a rate of 327 W.

Part A

What is its resistance?

Part B

What is the current in the resistor?

### Answers

The resistor has a **resistance **of 0.688. 21.8 A of current flow via the resistor.

What is the difference in potential between a resistor's ends?

Resistance is the name given to the **proportional constant**. We now know that resistance rises as temperature rises. So, if a resistor's temperature is constant, we may deduce that the potential difference at its ends is directly proportional to the current flowing through it.

R = V²/P

R = (15.0 V)²/327 W = 0.688 Ω

Using Ohm's Law, which states that V = IR, where V is the potential difference, I is the **current**, and R is the resistance, we can rearrange to solve for I:

I = V/R

I = 15.0 V/0.688 Ω = 21.8 A

Therefore, the current in the resistor is 21.8 A.

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a particle with charge q = –1 c is moving in the positive z-direction at 5 m/s. the magnetic field at its position is B = (3 i- 4 j)T what is the magnetic force on the particle?A. (20ỉ +15j) N B. (20î–153) N C. (-20i +15j) N D.(-20i -15j) N E. none of these

### Answers

The magnetic force on the **negatively charged particle** moving in the positive z-direction with velocity 5 m/s in a magnetic field of (3 i- 4 j) T is (-20i -15j) N.

How to find magnetic force on the particle?

The **magnetic force** on a particle with charge q moving at velocity v in a magnetic field B is given by the equation F = q(v × B), where × represents the vector cross product.

In this case, the particle has a charge of q = -1 C and is moving in the positive **z-direction** at 5 m/s, so its velocity is given by v = (0, 0, 5) m/s.

The magnetic field at its position is B = (3, -4, 0) T.

Taking the vector** cross product** of v and B, we get:

v × B = (5, 0, 0) × (3, -4, 0) = (0, 0, -20)

So the magnetic force on the particle is given by:

F = q(v × B) = -1 C × (0, 0, -20) N = (0, 0, 20) N

Therefore, the correct answer is (D) (-20i -15j) N.

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the horizontal surface on which the objects slide is frictionless if fl = 29n and fr = 3n what is the magnitude of the force exerted on the block with mass 7kg by the block with mass 5kg

### Answers

The total amount of force exerted on an object is referred to as the **magnitude of force**. The strength of the force increases when all the forces are pulling in the same direction. When forces are exerted on an item from different angles, the force's strength reduces.

To solve this problem, we need to use** Newton's Second Law**, which states that the force exerted on an object is equal to its mass multiplied by its acceleration. In this case, we want to find the force exerted on the 7kg block by the 5 kg block.

First, we need to calculate the acceleration of the system. Since the surface is frictionless, there is no force opposing the motion of the blocks in the horizontal direction. Therefore, the net force in the horizontal direction is simply the force exerted by the 5kg block, which we can calculate using the formula: F = ma

where F is the net force, m is the mass of the block (5 kg), and a is the acceleration of the system.

F = ma

F = 5 kg x a

We can rearrange this equation to solve for the acceleration:

a = F/m

a = 29N/5 kg

a = 5.8 m/s^2

Now that we know the acceleration of the system, we can use **Newton's Second Law **to find the force exerted on the 7kg block:

F = ma

F = 7 kg x 5.8 m/s^2

F = 40.6 N

Therefore, the **magnitude of the force** exerted on the 7kg block by the 5 kg block is 40.6 N.

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Water flows in a 150-mm diameter pipe at 5.5 m/s. Is this flow laminar or turbulent? 2. Oil with viscosity 50 mPa.s and density 900 kg/m3 flows along a 20 cm- diameter pipe. Find the maximum velocity in order to maintain laminar low.

### Answers

The given flow is **turbulent **and the maximum velocity of laminar flow is **0.64 m/s **

1. To determine if the flow is laminar or turbulent, we can use the **Reynolds number** formula:

Re = (ρVD)/μ

Where:

ρ = density of fluid

V = velocity of fluid

D = diameter of pipe

μ = dynamic viscosity of fluid

Plugging in the values given, we get:

Re = (1000 kg/m3)(5.5 m/s)(0.15 m)/(0.001 kg/m.s) = 90750

If the Reynolds number is less than 2300, the flow is laminar. If it is greater than 4000, the flow is turbulent. If it is between 2300 and 4000, the flow may be laminar or turbulent depending on other factors.

In this case, the Reynolds number is greater than 4000, so the flow is **turbulent**.

2. To maintain laminar flow, the Reynolds number should be less than 2300. We can rearrange the Reynolds number formula to solve for the maximum velocity:

Vmax = (Reμ)/(ρD)

Plugging in the values given, we get:

Vmax = (2300)(0.05 Pa.s)/(900 kg/m3)(0.2 m) = 0.64 m/s

Therefore, the **maximum velocity** to maintain laminar flow in this pipe is **0.64 m/s**.

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The voltage transfer function for a first-order circuit is T(s) = 5s/(s + 15,000).

Find the passband gain and the cutoff frequency. b)Use MATLAB to plot the magnitude of the Bode gain response.

### Answers

The** passband gain **for the given first-order circuit is 5, and the cutoff frequency is 15,000 rad/s.

To find the passband gain and **cutoff frequency** of the voltage transfer function T(s) = 5s/(s + 15,000), follow these steps:

1. Passband gain: In a first-order circuit, the passband gain is simply the coefficient of 's' in the numerator, which is 5 in this case.

2. Cutoff frequency: The cutoff frequency is the value of 's' when the **denominator **is equal to the numerator. In this case, set 5s = s + 15,000, which gives s = 15,000 rad/s as the cutoff frequency.

To plot the Bode gain response using **MATLAB**, use the following code:

```MATLAB

numerator = [5 0];

denominator = [1 15000];

sys = tff(numerator, denominator);

bode(sys);

```

This code creates a transfer function object 'sys' with the given numerator and denominator coefficients, then uses the 'bode()' function to plot the **Bode gain **response.

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A conducting disk with radius a, height h 《 a, and conductivity σ is immersed in a time varying but spatially uniform magnetic field parallel to its axis. B(t) = Bo sin(wt)2 a) [6 pts] Ignoring the effects of any induced magnetic fields, find the induced electric field E(r, t) and the current density J(r, t) in the disk. Sketch the current distribution. b) [3 pts If the power dissipated in a resistor is P IV, show that the power dissipated per unit volume is J-E. c) [3 pts] Use your results from parts a) and b) to calculate the total power dissipated in the disk at time t, and the average power dissipated per cycle of the field. d) [6 pts] Suppose the disk was roughly the size of the solid base of a typical frying pan, and the frequency was 20 kHz, what approximate scale for Bo would you need to significantly heat up the pan (say, 1000 watts of power). Does this seem feasible? Then, use the current distribution in part a) to determine the induced magnetic field at the center of the pan. Is the induced magnetic field small compared to the applied field?

### Answers

a) The induced electric field E(r, t) in the disk can be obtained from Faraday's** law of induction,** which states that the electromotive force (EMF) around a closed loop is equal to the negative time derivative of the magnetic flux through the loop:

EMF = -dΦ/dt

For a **conducting disk, **the EMF is related to the induced electric field E(r, t) and the circumference of the disk C by:

EMF = ∮ E(r,t)[tex]· dl = E(r,t) C[/tex]

where the integral is taken over the circumference of the disk. The magnetic flux Φ through the disk can be calculated from the magnetic field B(t) and the surface area of the disk A = πa^2:

Φ = ∫ B(t) · dA = B(t) πa^2

Taking the time derivative of Φ, we get:

[tex]dΦ/dt = πa^2 d/dt (B(t) sin^2(wt))\\= 2πa^2 B(t) w cos(wt) sin(wt)[/tex]

Therefore, the **induced** electric field in the disk is:

E(r, t) = -dΦ/dt / C

= -2a B(t) w cos(wt) sin(wt)

The current density J(r, t) can be obtained from Ohm's law, which relates the current density to the electric field and the conductivity σ:

J(r, t) = σ E(r, t)

= -2a σ B(t) w cos(wt) sin(wt)

The current density is proportional to the sine of the azimuthal angle φ in **cylindrical coordinates**, and has a maximum value at the edge of the disk (r = a).

The arrows indicate the direction of the current density, which flows clockwise around the disk.

b) The power dissipated per unit volume is given by the product of the current density and the electric field:

P/V = J · E

= [tex]4a^2 σ B^2 w^2 sin^2(wt)[/tex]

c) The total power dissipated in the disk at time t is obtained by integrating the power density over the volume of the disk:

P(t) = ∫ P/V · dV

= [tex]∫0^h ∫0^a 4a^2 σ B^2 w^2 sin^2(wt) · r dr dz dφ\\= 2πa^3h σ B^2 w^2 sin^2(wt)[/tex]

The average **power** dissipated per cycle of the field is one half of the maximum power dissipated:

[tex]P_avg = (1/2) · (2πa^3h σ B^2 w^2 / 2)\\= πa^3h σ B^2 w^2[/tex]

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Three liquids are at temperatures of 5◦C, 19◦C, and 39◦C, respectively. Equal masses of the first two liquids are mixed, and the equi- librium temperature is 13◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 33.2◦C.

Find the equilibrium temperature when equal masses of the first and third are mixed.

Answer in units of ◦C.

### Answers

When equal masses of the first (5°C) and third (39°C) liquids are mixed, the equilibrium temperature will be 22°C. This can be found using the equation avgT = (m1*T1 + m2*T2)/(m1+m2), where avgT is the average temperature, m1 and m2 are the masses of the first and second liquids respectively, and T1 and T2 are the temperatures of the first and second liquids respectively.

In the first scenario, we are mixing equal masses of the first (5°C) and second (19°C) liquids. Therefore, m1 = m2 = m, and T1 = 5°C and T2 = 19°C. Plugging this into the equation yields avgT = (m*5 + m*19)/(m+m), which simplifies to avgT = (5m + 19m)/2m = 24m/2m = 24/2 = 12°C. Since this is the equilibrium temperature, this means that it is also equal to the final temperature of the mixture, which is 13°C.

In the second scenario, we are mixing equal masses of the second (19°C) and third (39°C) liquids. Therefore, m1 = m2 = m, and T1 = 19°C and T2 = 39°C. Plugging this into the equation yields avgT = (m*19 + m*39)/(m+m), which simplifies to avgT = (19m + 39m)/2m = 58m/2m = 58/2 = 29°C. Since this is also the equilibrium temperature, this also means that it is equal to the final temperature of the mixture, which is 33.2°C.

Using the same equation and the values for the first and third liquids (5°C and 39°C), we can find the equilibrium temperature for when equal masses of the first and third are mixed. Plugging the values into the equation yields avgT = (m*5 + m*39)/(m+m), which simplifies to avgT = (5m + 39m)/2m = 44m/2m = 44/2 = 22°C. This is the equilibrium temperature when equal masses of the first and third liquids are mixed.

Determine the magnitude and direction of the force between two parallel wires 30 m long and 6.0 cm apart, each carrying 30A in the same direction. Determine the magnitude of the force between two parallel wires.

### Answers

The magnitude of the** force **between two parallel wires 30 m long and 6.0 cm apart, each carrying 30A in the same direction is** 0.03 N**. The magnitude of the force between two parallel wires can be calculated using the formula:

F = (μ₀ * I₁ * I₂ * L) / (2πd)

where F is the magnitude of the force, μ₀ is the **permeability constant** (4π x 10 N/A), I₁ and I₂ are the **currents** flowing through the wires, L is the **length of the wires**, and d is the **distance **between the wires.

Substituting the given values, we get:

F = (4π x 10 N/A * 30A * 30A * 30m) / (2π * 0.06m)

F = 0.03 N

The** **direction of the force can be determined using the** right-hand rule**. If we point the thumb of our right hand in the direction of the current in one wire and the fingers in the direction of the** current** in the other wire, the direction of the **force** on the wires is given by the direction of the palm.

In this case, since the currents are flowing in the same direction, the force between the wires is** attractive**, pulling the wires towards each other. Therefore, the magnitude of the force between two parallel wires 30 m long and 6.0 cm apart, each carrying 30A in the same direction is 0.03 N and the direction of the force is attractive, pulling the wires towards each other.

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6. The electric field has a magnitude of 3. 0 N/C at a distance of 30 cm from a point charge. What is the charge?

A) 1. 4 nC B) 30 pC C) 36 mC D) 12 mC

### Answers

The charge is 30 pC, when** electric field **has a magnitude of 3.0 N/C at the distance of 30 cm from point charge. Option B is correct.

We will use **Coulomb's law** to solve this problem;

Coulomb's law states that the magnitude of the electric field E created by a point charge Q at a distance r from the charge is given by;

E = k × Q / r²

where k is Coulomb's constant, which is approximately 9 x 10⁹ Nm²/C².

In this problem, we are given electric field having a magnitude of 3.0 N/C at a distance of 30 cm from a** point charge**. Converting 30 cm to meters, we have;

r = 0.3 m

Plugging the given values into Coulomb's law, we have;

3.0 N/C = (9 x 10⁹ Nm²/C²) × Q / (0.3 m)²

Solving for Q, we get;

Q = (3.0 N/C) × (0.3 m)² / (9 x 10⁹ Nm²/C²)

Q = 30 x 10⁻¹² C

Q = 30 pC

Therefore, the **charge** is 30 pC.

Hence, B. is the correct option.

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uppose that a civilization around a nearby star had television like we do. could current seti efforts detect their television transmissions? why or why not?

### Answers

**Answer: **Current **SETI** investigations are unlikely to catch television signals from a nearby civilisation because they are too faint to detect at stellar distances.

**Explanation: **Television signals are weaker than radio signals because they are sent in a narrow beam aimed at Earth's surface. These signals diminish quickly in space, making them hard to detect at enormous distances between stars. **Radio emissions**, which may be detected at interstellar distances, are the main SETI emphasis.

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Calculate the time required to fly from P to B, in terms of the eccentricity e and the period T. B lies on the minor axis.

### Answers

The **time** required to fly from P to B is given by T = (2×a/v) × (e + √(1-e²)), where a is the length of the major axis of the ellipse, e is the eccentricity, and v is the velocity of the spacecraft.

What is Kepler's second law?

Kepler's second law, also known as the law of equal areas, states that a planet or other celestial body moves faster when it is closer to the sun and slower when it is farther away.

Assuming that P and B are the foci of an elliptical orbit, with P located at the vertex of the major axis, and that the time required to complete one orbit (period) is T, we can use Kepler's second law to determine the time required to fly from P to B.

Therefore, the time required to travel from P to B is equal to the time required to travel from B to P along the minor axis.

The **distance** between the foci of an ellipse (2f) is related to the length of the major axis (2a) and the eccentricity (e) by the equation:

2f = 2a×e

Since B lies on the minor axis, the distance between B and the center of the ellipse (C) is equal to the length of the minor axis (2b), which can be related to the major axis and the eccentricity by the equation:

2b = 2a×√(1-e²)

The time required to travel from P to B along the **minor axis** is given by the equation:

T/2 = (1/2) × [(2b + 2f) / v]

Substituting the expressions for 2f and 2b gives:

T/2 = (1/2) × [(2ae + 2a√(1-e²)) / v]

Simplifying the expression gives:

T = (2×a/v) × (e + √(1-e²))

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