Two small charged objects repel each other with a force of 32 N when they are separated by a distance d. If the charge on each object is reduced by one third of its original value and the distance between them is doubled, then the new force between them is?

The transformation equation to convert Celsius temperatures (C) to Joe Scientist's temperature scale (J) is:

J = 2.39C + 57

In Joe Scientist's temperature scale,

water freezes = 57

water   boils =  296.

In Celsius scale, water freezes at 0 and boils at 100.

To convert Celsius temperatures (C) to Joe Scientist's scale temperatures (J), we can use a linear transformation equation.

The general equation for linear transformation is:

J = aC + b

Celsius: 0 (water freezing point) -> Joe Scientist: 57

Celsius: 100 (water boiling point) -> Joe Scientist: 296

we can set up a system of linear equations to solve for 'a' and 'b' provided we have  the data points

Equation 1: 0a + b = 57

Equation 2: 100a + b = 296

We solve this and find that

'a' =2.39

'b'=  57.

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The magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).

The magnetic flux through a coil can be calculated using the formula:

Φ = L * I

where Φ is the magnetic flux, L is the inductance of the coil, and I is the current passing through the coil.

Given:

Number of turns in the coil (N) = 420

Inductance of the coil (L) = 11 mH = 11 × 10^(-3) H

Current passing through the coil (I) = 4.7 mA = 4.7 × 10^(-3) A

First, we need to calculate the effective number of turns by multiplying the number of turns with the current:

[tex]N_eff = N * IN_eff = 420 * 4.7 × 10^(-3)N_eff = 1.974\\[/tex]

Now, we can calculate the magnetic flux using the formula:

[tex]Φ = L * IΦ = (11 × 10^(-3) H) * (1.974)Φ = 21.714 × 10^(-3) WbΦ = 21.714 mWb\\[/tex]

Therefore, the magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).

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Answer: It would be A. Impedance of the circuit

Explanation:

As the distance between two positive charges is increased, the potential energy of the system decreases.

The potential energy between two charges is given by the equation U = k * (q1 * q2) / r, where U is the potential energy, k is the electrostatic constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.Since the charges are positive, their potential energy is positive as well. As the distance between the charges increases (r increases), the denominator of the equation gets larger, resulting in a smaller potential energy. Therefore, the potential energy decreases as the distance between the charges is increased. In summary, the potential energy decreases as the distance between two positive charges is increased.

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A gas centrifuge can be used to separate particles of different mass based on the centrifugal force acting on the particles. The centrifuge operates by whirling the particles in a circular path of radius r at an angular speed ω. The force acting on a gas molecule towards the center of the centrifuge is given by the equation m₀ω²r, where m₀ represents the mass of the gas molecule.

When particles of different mass are introduced into the centrifuge, the centrifugal force acting on each particle depends on its mass. Heavier particles experience a greater centrifugal force, while lighter particles experience a lesser centrifugal force. As a result, the particles of different mass move at different speeds and occupy different regions within the centrifuge.

Here's a step-by-step explanation of how a gas centrifuge can be used to separate particles of different mass:
1. Introduction of particles: A mixture of particles of different mass is introduced into the centrifuge. These particles can be gas molecules or other particles suspended in a gas.
2. Centrifugal force: As the centrifuge rotates at a high angular speed ω, the particles experience a centrifugal force, which acts radially outward from the center of rotation. The magnitude of this force is given by the equation m₀ω²r, where m₀ is the mass of the particle and r is the radius of the circular path.
3. Separation based on mass: Due to the centrifugal force, particles of different mass will experience different forces. Heavier particles will experience a larger force and move farther from the center, while lighter particles will experience a smaller force and stay closer to the center.
4. Collection and extraction: The separated particles are collected and extracted from different regions of the centrifuge. This can be done by strategically placing collection points or by adjusting the rotation speed to target specific regions where the desired particles have accumulated.

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The intensity of the light after passing through the polarizing filter with an axis making an angle of 07 degrees from the horizontal is approximately 155 W/m².

When light passes through a polarizing filter, the intensity of the transmitted light is given by Malus's law:

I = I₀ * cos²(θ)

Where:

I₀ = initial intensity of the light

θ = angle between the polarization axis of the filter and the direction of polarization of the incident light

I = intensity of the transmitted light

Given:

Initial intensity (I₀) = 160 W/m²

Angle (θ) = 07 degrees

Converting the angle to radians:

θ = 07 degrees * (π/180) ≈ 0.122 radians

Applying Malus's law:

I = I₀ * cos²(θ)

I = 160 W/m² * cos²(0.122)

Calculating the intensity:

I ≈ 160 W/m² * cos²(0.122)

I ≈ 160 W/m² * 0.973

Expressing the intensity with the appropriate units:

I ≈ 155 W/m²

Therefore, the intensity of the light after passing through the polarizing filter with an axis making an angle of 07 degrees from the horizontal is approximately 155 W/m².

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The net work done by the net torque on the ball to make it come to rest is approximately -8.422 Joules.

To find the net work done by the net torque on the ball to make it come to rest, we need to use the rotational kinetic energy equation:

K_rot = (1/2) * I * ω²

Where:

K_rot is the rotational kinetic energy

I is the moment of inertia of the ball

ω is the angular velocity

The moment of inertia of a solid sphere rotating about its axis of symmetry can be calculated using the formula:

I = (2/5) * m * r²

Where:

m is the mass of the ball

r is the radius of the ball

Given:

Mass of the ball (m) = 2.860 kg

Diameter of the ball = 60.000 cm

Angular velocity (ω) = 5.100 rev/s

First, we need to convert the diameter of the ball to its radius:

Radius (r) = Diameter / 2 = 60.000 cm / 2 = 30.000 cm = 0.300 m

Now, we can calculate the moment of inertia (I) using the formula:

I = (2/5) * m * r² = (2/5) * 2.860 kg * (0.300 m)²

I = 0.3432 kg·m²

Next, we can calculate the initial rotational kinetic energy (K_rot_initial) using the given angular velocity:

K_rot_initial = (1/2) * I * ω² = (1/2) * 0.3432 kg·m² * (5.100 rev/s)²

K_rot_initial = 8.422 J

Since the net torque causes the ball to come to rest, the final rotational kinetic energy (K_rot_final) is zero. The net work done by the net torque can be calculated as the change in rotational kinetic energy:

Net Work = K_rot_final - K_rot_initial = 0 - 8.422 J

Net Work = -8.422 J

Therefore, the net work done by the net torque on the ball to make it come to rest is approximately -8.422 Joules (to three decimal places).

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The difference in rotational kinetic energy when the ice dancer pulls in her arms from a moment of inertia of 2.74 kg m² to 1.54 kg m² is 0.998 Joules.

When the ice dancer pulls in her arms, her moment of inertia decreases, resulting in a change in rotational kinetic energy. The formula for the difference in rotational kinetic energy (ΔK) is given by ΔK = ½ * (I₂ - I₁) * (ω₂² - ω₁²), where I₁ and I₂ are the initial and final moments of inertia, and ω₁ and ω₂ are the initial and final angular velocities.

Given I₁ = 2.74 kg m², I₂ = 1.54 kg m², and ω₁ = 2.2 rad/s, we can calculate ω₂ using the conservation of angular momentum, I₁ * ω₁ = I₂ * ω₂. Solving for ω₂ gives ω₂ = (I₁ * ω₁) / I₂.

Substituting the values into the formula for ΔK, we have ΔK = ½ * (I₂ - I₁) * [(I₁ * ω₁ / I₂)² - ω₁²].

Performing the calculations, we find ΔK ≈ 0.998 Joules. This means that when the ice dancer pulls in her arms, the rotational kinetic energy decreases by approximately 0.998 Joules.

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The currents I /,I 2 and I 3 in the circuit using Kirchhoff's Rules is  0.16 A.

Kirchhoff’s Rules are used to explain the distribution of electric current in circuits, and to calculate the potential difference between any two points on a circuit. In the given circuit, the first step is to identify the junctions and branches, there are two junctions, namely J1 and J2, and three branches, which are B1, B2, and B3. Once these have been identified, it is possible to use Kirchhoff's Rules to determine the currents. First, apply Kirchhoff's first law at junction J1, the total current entering the junction must equal the total current leaving the junction.

Therefore:I1 = I2 + I3 Second, apply Kirchhoff's second law in each of the loops.

For example, for loop 1-2-3-4-1:−4V + 10Ω(I1 − I2) + 20Ω(I1 − I3) = 0

Using Kirchhoff's second law on all three loops gives the following system of equations:10I1 − 10I2 − 20I3 = 4−10I1 + 30I2 − 10I3 = 0−20I1 − 10I2 + 30I3 = 0

Solving this system of equations gives I1 = 0.24 A, I2 = 0.18 A, and I3 = 0.16 A. Therefore, the currents are:I1 = 0.24 AI2 = 0.18 AI3 = 0.16 A.

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(a) Length of the pendulum(l) = 0.64 m which can be calculated by using the formula, T = 2π√(l/g) where T = time period. we have to use the length and acceleration due to gravity.

The angle of a pendulum as a function of time is given as (t) = (0.19 rad)cos((3.9 Hz)t + 0.48 rad). The length of the pendulum can be determined by using the formula, T = 2π√(l/g) where T = time period, g = acceleration due to gravity = 9.81 m/s², and l = length of the pendulum.

Since the time period of the given pendulum is not given directly, we can find it by converting the given frequency into the time period. Frequency(f) = 3.9 Hz

Time period(T) = 1/f = 1/3.9 s= 0.2564 s

Now, substituting the value of time period and acceleration due to gravity in the above formula, we have;`T = 2π√(l/g) 0.2564 = 2π√(l/9.81)

On solving the above equation, we get;

l = (0.2564/2π)² × 9.81

Length of the pendulum(l) = 0.64 m

(b) The amplitude of the pendulum's motion is 10.89° which will be obtained from the equation Angle(t) = 0.19 cos(3.9t) + 0.48 rad.

The amplitude of the given pendulum can be determined as follows; Angle(t) = 0.19 cos(3.9t) + 0.48 rad

Comparing it with the standard equation of the cosine function, we can say that the amplitude of the given pendulum is 0.19 rad or 10.89°. Hence, the amplitude of the pendulum's motion is 10.89°.

(c) Determine the period of the pendulum's motion, in s.

The period of the pendulum's motion is 0.256 s.

The period of the given pendulum can be determined using the following formula, T = 2π/ω where T = time period, and ω = angular frequency. Since the value of the angular frequency is not given directly, we can obtain it from the given frequency.`Frequency(f) = 3.9 Hz`Angular frequency(ω) = 2πf= 2π × 3.9= 24.52 rad/s

Now, substituting the value of angular frequency in the above formula, we have; T = 2π/ω`= `2π/24.52`= `0.256` s

Hence, the period of the pendulum's motion is 0.256 s.

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The half-life of the radioisotope is 30 minutes. The half-life of a radioisotope is the time it takes for half of the nuclei in a sample to decay.

In this case, we start with 400 nuclei and after one hour, only 25 nuclei remain. This means that 375 nuclei have decayed in one hour. Since the half-life is the time it takes for half of the nuclei to decay, we can calculate it by dividing the total time (one hour or 60 minutes) by the number of times the half-life fits into the total time.

In this case, if 375 nuclei have decayed in one hour, that represents half of the initial sample size (400/2 = 200 nuclei). Therefore, the half-life is 60 minutes divided by the number of times the half-life fits into the total time, which is 60 minutes divided by the number of half-lives that have occurred (375/200 = 1.875).

Therefore, the half-life of the isotope is approximately 30 minutes.

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The frequency of revolution of the electron is approximately 1.92x10¹⁴ Hz. The radius of the circular path of the electron is approximately 5.61x10⁻³ m.

To solve this problem, we can use the equation for the frequency of revolution of a charged particle in a magnetic field:

(a) The frequency of revolution, f, is given by the equation:

f = qB / (2πm)

f is the frequency of revolution

q is the charge of the electron (1.6x10⁻¹⁹ C)

B is the magnitude of the magnetic field (70 μT = 70x10⁻⁶ T)

m is the mass of the electron (9.11x10⁻³¹ kg)

Let's plug in the values:

f = (1.6x10⁻¹⁹ C)(70x10⁻⁶ T) / (2π)(9.11x10⁻³¹kg)

Calculating this expression gives:

f ≈ 1.92x10¹⁴ Hz

So, the frequency of revolution of the electron is approximately 1.92x10¹⁴ Hz.

(b) The radius of the circular path of the electron, r, can be determined using the equation for the centripetal force:

F = qvB = mv² / r

F is the force acting on the electron due to the magnetic field

v is the velocity of the electron

Since the electron is moving in a circular path, we can equate the centripetal force to the magnetic force:

qvB = mv² / r

Simplifying and solving for r, we get:

r = mv / (qB)

Let's calculate the radius using the given values:

r = (9.11x10⁻³¹ kg)(√(2(189 eV)(1.6x10⁻¹⁹ J/eV))) / ((1.6x10⁻¹⁹ C)(70x10⁻⁶ T))

Calculating this expression gives:

r ≈ 5.61x10⁻³ m

Therefore, the radius of the circular path of the electron is approximately 5.61x10⁻³ m.

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The resistance of the light bulb can be determined using Ohm's Law, which states that the resistance (R) is equal to the ratio of the potential difference (V) across the bulb to the current (I) passing through it:

R = V / I

Given:

Potential difference (V) = 120 V

Current (I) = 0.83 A

Substituting these values into the formula:

R = 120 V / 0.83 A

R ≈ 144.58 Ω (rounded to two decimal places)

Therefore, the resistance of the light bulb is approximately 144.58 Ω.

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Answer:

a. To determine the energy of the photon needed to excite an electron from the b level to the e level in the Mercury atom, you would need to know the specific energy values for each level. Typically, energy levels are represented in electron volts (eV) or joules (J) in atomic spectroscopy.

b. Once you have determined the energy difference between the b and e levels, you can convert it to joules using the conversion factor 1 eV = 1.602 x 10^(-19) J.

c. The frequency of a photon can be calculated using the equation E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^(-34) J·s), and f is the frequency. Rearranging the equation, you can solve for f: f = E / h.

d. The color of the emitted photon is determined by its wavelength or frequency. The relationship between wavelength (λ) and frequency (f) is given by the equation c = λf, where c is the speed of light (~3 x 10^8 m/s). Different wavelengths correspond to different colors in the electromagnetic spectrum. You can use this relationship to determine the color of the photon once you have its frequency or wavelength.

To obtain specific values for the energy levels, you may need to refer to a reliable reference source or consult a physics or atomic spectroscopy textbook.

The cooling rate of the object is 0.054.

Let's find the cooling rate (k) of an object using the given information. Ts = 10 °CTo = 110 °CT1 = 35 °Ct2 = 25 minutes. Now, the given formula is T = Ts + (To - Ts) e ^ -kt. Here, we know that the temperature drops from 110°C to 35°C, which is 75°C in 25 minutes. Now, we will substitute the values in the formula as follows:35 = 10 + (110 - 10) e ^ (-k × 25) => (35 - 10) / 100 = e ^ (-k × 25) => 25 / 100 = k × 25 => k = 0.054. Therefore, the cooling rate of the object is 0.054. Hence, option A is correct.

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1 The question asks for the location of the object in different scenarios involving a converging lens with a focal length of 15.9 cm. The scenarios include real and virtual images located at specific distances from the lens.

In scenario (a), where a real image is located at a distance of 47.7 cm from the lens, we can use the lens formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. Rearranging the formula, we get 1/u = 1/f - 1/v. Plugging in the given values, we have 1/u = 1/15.9 - 1/47.7. Solving this equation gives us the object distance u.

In scenario (b), the real image is located at a distance of 95.4 cm from the lens. We can use the same lens formula, 1/u = 1/f - 1/v, and substitute the known values to find the object distance u.

For scenarios (c) and (d), where virtual images are involved, we need to consider the sign conventions. A negative sign indicates that the image is virtual. Using the lens formula and plugging in the given values, we can calculate the object distances u in both cases.

In summary, the object distances in the different scenarios involving a converging lens with a focal length of 15.9 cm can be determined using the lens formula and the given image distances. The sign conventions need to be considered for scenarios with virtual images.Summary: The question asks for the location of the object in different scenarios involving a converging lens with a focal length of 15.9 cm. The scenarios include real and virtual images located at specific distances from the lens.

In scenario (a), where a real image is located at a distance of 47.7 cm from the lens, we can use the lens formula, 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. Rearranging the formula, we get 1/u = 1/f - 1/v. Plugging in the given values, we have 1/u = 1/15.9 - 1/47.7. Solving this equation gives us the object distance u.

In scenario (b), the real image is located at a distance of 95.4 cm from the lens. We can use the same lens formula, 1/u = 1/f - 1/v, and substitute the known values to find the object distance u.

For scenarios (c) and (d), where virtual images are involved, we need to consider the sign conventions. A negative sign indicates that the image is virtual. Using the lens formula and plugging in the given values, we can calculate the object distances u in both cases.

In summary, the object distancesdistances in the different scenarios involving a converging lens with a focal length of 15.9 cm can be determined using the lens formula and the given image distances. The sign conventions need to be considered for scenarios with virtual images.

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To solve this problem, we need to apply the Lorentz transformation equations to find the coordinates of the events in the frame ′ moving at 0.70c relative to the observer's frame.

The Lorentz transformation equations are as follows:

x' = γ(x - vt)

y' = y

z' = z

t' = γ(t - vx/c^2)

where γ is the Lorentz factor, v is the relative velocity between the frames, c is the speed of light, x, y, z, and t are the coordinates in the observer's frame, and x', y', z', and t' are the coordinates in the moving frame ′.

Given:

x1 = y1 = z1 = t1 = 0

x2 = 200 m, y2 = z2 = 0

(a) To find the coordinates of the events in the frame ′, we substitute the given values into the Lorentz transformation equations. Since y and z remain unchanged, we only need to calculate x' and t':

For the first event:

x'1 = γ(x1 - vt1)

t'1 = γ(t1 - vx1/c^2)

Substituting the given values and using v = 0.70c, we have:

x'1 = γ(0 - 0)

t'1 = γ(0 - 0)

For the second event:

x'2 = γ(x2 - vt2)

t'2 = γ(t2 - vx2/c^2)

Substituting the given values, we get:

x'2 = γ(200 - 0.70c * t2)

t'2 = γ(t2 - 0.70c * x2/c^2)

(b) The distance between the events in the frame ′ is given by the difference in the transformed x-coordinates:

Δx' = x'2 - x'1

(c) To determine if the events are simultaneous in the frame ′, we compare the transformed t-coordinates:

Δt' = t'2 - t'1

Now, let's calculate the values:

(a) For the first event:

x'1 = γ(0 - 0) = 0

t'1 = γ(0 - 0) = 0

For the second event:

x'2 = γ(200 - 0.70c * t2)

t'2 = γ(t2 - 0.70c * x2/c^2)

(b) The distance between the events in the frame ′ is given by:

Δx' = x'2 - x'1 = γ(200 - 0.70c * t2) - 0

(c) To determine if the events are simultaneous in the frame ′, we calculate:

Δt' = t'2 - t'1 = γ(t2 - 0.70c * x2/c^2) - 0

In order to proceed with the calculations, we need to know the value of the relative velocity v.

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a. The maximum velocity of the object in m/s if the object bounces 2.95 cm above and below its equilibrium position is sqrt((1.20 N/m * (0.0295 m)^2) / 0.0170 kg).

b.  The maximum velocity of the object is done

(maximum velocity)^2

(a) To determine the maximum velocity of the object, we can use the principle of conservation of mechanical energy. At the maximum displacement, all of the potential energy is converted into kinetic energy.

The potential energy (PE) of the object can be calculated using the formula:

PE = 0.5 * k * x^2

where k is the force constant of the spring and x is the displacement from the equilibrium position.

Mass of the object (m) = 0.0170 kg

Force constant of the spring (k) = 1.20 N/m

Displacement from equilibrium (x) = 2.95 cm = 0.0295 m

The potential energy can be calculated as follows:

[tex]PE = 0.5 * k * x^2 = 0.5 * 1.20 N/m * (0.0295 m)^2[/tex]

To find the maximum velocity, we equate the potential energy to the kinetic energy (KE) at the maximum displacement:

PE = KE

[tex]0.5 * 1.20 N/m * (0.0295 m)^2 = 0.5 * m * v^2[/tex]

Simplifying the equation and solving for v:

[tex]v = sqrt((k * x^2) / m[/tex]

[tex]v = sqrt((1.20 N/m * (0.0295 m)^2) / 0.0170 kg)[/tex]

Calculating this expression will give us the maximum velocity of the object in m/s.

(b) The kinetic energy (KE) at the maximum velocity can be calculated using the formula:

[tex]KE = 0.5 * m * v^2[/tex]

Mass of the object (m) = 0.0170 kg

Maximum velocity (v) = the value calculated in part (a)

Plugging in the values, we can calculate the kinetic energy in Joules.

[tex]KE = 0.5 * 0.0170 kg *[/tex] (maximum velocity)^2

Calculating this expression will give us the Joules of kinetic energy at the maximum velocity.

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According to the law of conservation of energy, the sum of kinetic energy and potential energy remains constant for a system. Therefore, any gain or loss in potential energy will lead to an equal and opposite change in kinetic energy. As a result, the total energy of the system is conserved.

Two balls, 1 and 2, of equal mass and radius, each rotate around their fixed central axis. If ball 1 rotates with an angular speed equal to three times the angular speed of ball 2, find the ratio KE:/KE. As given, both balls have the same mass and radius. Therefore, they have the same moment of inertia. The moment of inertia of a sphere rotating about its diameter is given by,I = (2/5) MR²Since both the balls have the same mass and radius, they will have the same moment of inertia.I₁ = I₂ = (2/5) MR².

Now, let the angular speed of ball 2 be ω rad/s. Therefore, the angular speed of ball 1 is 3ω rad/s. Both the balls have the same moment of inertia, so the rotational kinetic energy of each ball will be the same. It is given by,KER = (1/2) I ω²Therefore,KER₁ = KER₂ = (1/2) I ω² = (1/2) (2/5) MR² ω² = (1/5) MR² ω²Now, let's calculate the ratio KE₁ / KE₂.KE₁ / KE₂ = KER₁ / KER₂= [(1/5) MR² ω₁²] / [(1/5) MR² ω₂²]= ω₁² / ω₂²= (3ω₂)² / ω₂²= 9ω₂² / ω₂²= 9/1= 9:1Therefore, the required ratio KE₁ / KE₂ is 9:1.

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The object's charge is -4.5x10^-5 C.

we can use the formula for electric field strength (E) due to a point charge:

E = k * (|Q| / r^2)

Where:

E = Electric field strength

k = Coulomb's constant (8.99x10^9 N m^2/C^2)

|Q| = Absolute value of the charge on the object

r = Distance from the object

Rearranging the formula, we can solve for |Q|:

|Q| = E * (r^2 / k)

Plugging in the given values:

E = 2.5x10^5 N/C

r = 1.8 cm = 0.018 m

k = 8.99x10^9 N m^2/C^2

|Q| = (2.5x10^5 N/C) * (0.018 m)^2 / (8.99x10^9 N m^2/C^2)

= 4.5x10^-5 C

Since the electric field points away from the object, the charge must be negative, so the object's charge is approximately -4.5x10^-5 C.

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The magnitude and direction of the electric field due to the three charges at the vacant corner can be calculated using Coulomb's law. Coulomb's law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.The electric field at the vacant corner is the vector sum of the electric fields due to the other three charges.

The magnitude of the electric field due to each of the three charges is given by;E = kq / r²where k is the Coulomb constant, q is the charge, and r is the distance between the charges.The distance between each of the charges and the vacant corner can be calculated using the Pythagorean theorem since they are placed at the three corners of a square 40mm on a side.

Thus, the distance between each charge and the vacant corner is:√(40² + 40²) = 56.6 mmThe magnitude of the electric field due to each of the charges is:

E = (9 x 10⁹) x (6 x 10⁻⁹) / (0.0566)²E

= 45.4 N/C

The direction of the electric field due to the two charges on the horizontal side of the square will be at an angle of 45° to the x-axis, and the direction of the electric field due to the charge on the vertical side of the square will be at an angle of -45° to the y-axis.

Therefore, the resultant electric field at the vacant corner will be:E = √(45.4² + 45.4²) = 64.3 N/CThe angle made by the resultant electric field with the positive x-axis is given by:θ = tan⁻¹(45.4 / 45.4) = 45°Therefore, the magnitude and direction of the electric field due to the three charges at the vacant corner are 64.3 N/C and 45° with the positive x-axis, respectively.

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A merry-go-round accelerates from rest to 0.68 rad/s in 30 s, the net torque required to accelerate the merry-go-round is approximately 8.03×[tex]10^3[/tex] N·m.

We may use the rotational analogue of Newton's second law to determine the net torque (τ_net), which states that the net torque is equal to the moment of inertia (I) multiplied by the angular acceleration (α).

I = (1/2) * m * [tex]r^2[/tex]

I = (1/2) * (3.10×[tex]10^4[/tex] kg) * [tex](6.0 m)^2[/tex]

I ≈ 3.49×[tex]10^5[/tex] kg·[tex]m^2[/tex]

Now,

α = (ω_f - ω_i) / t

α = (0.68 rad/s - 0 rad/s) / (30 s)

α ≈ 0.023 rad/[tex]s^2[/tex]

So,

τ_net = I * α

Substituting the calculated values:

τ_net ≈ (3.49×[tex]10^5[/tex]) * (0.023)

τ_net ≈ 8.03×[tex]10^3[/tex] N·m

Therefore, the net torque required to accelerate the merry-go-round is approximately 8.03×[tex]10^3[/tex] N·m.

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The transmitted intensity through the three polarizing plates is approximately 1.296 units.

To determine the transmitted intensity through the three polarizing plates, considering Malus's Law,

I = Ii × cos²(θ)

Where:

I: transmitted intensity

Ii: incident intensity

θ: angle between the transmission axis of the polarizer and the plane of polarization of the incident light.

Given,  

Ii = 10.0 units  

θ1 = 20.0°

θ2 = 40.0°

θ3 = 60.0°

Calculate the transmitted intensity through each plate:

I₁ = 10.0 × cos²(20.0°)

I₁ ≈ 10.0 × (0.9397)²

I₁ ≈ 8.821 units

I₂ = 8.821 ×cos²(40.0°)

I₂ ≈ 8.821 ×(0.7660)²

I₂ ≈ 5.184 units

I₃ = 5.184 × cos²(60.0°)

I₃ ≈ 5.184 × (0.5000)²

I₃ ≈ 1.296 units

Therefore, the transmitted intensity is 1.296 units.

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(a) It will take 2 seconds for the ball to pass the man moving in the downward direction.

(b) The maximum height attained by the ball is 20 meters.

(c) The ball will take 2+2\sqrt{2} seconds to hit the ground.

(a) How long will it take the ball to pass the man moving in the downward direction?

We can use the equation of motion:

v = u + at,

where:

v = final velocity (0 m/s since the ball will momentarily stop when passing the man),

u = initial velocity (20 m/s upwards),

a = acceleration (due to gravity, -10 m/s²),

t = time.

Substituting the known values we get:

0 = 20 - 10t.

Simplifying the equation:

10t = 20,

t = 20/10,

t = 2 seconds.

Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.

(b) What is the maximum height attained by the ball?

To find the maximum height attained by the ball, we can use the following equation:

v² = u² + 2as,

where:

v = final velocity (0 m/s at the maximum height),

u = initial velocity (20 m/s upwards),

a = acceleration (acceleration due to gravity, -10 m/s²),

s = displacement.

The maximum height will be achieved when v = 0. Rearranging the equation, we get:

0 = (20)² + 2(-10)s.

Simplifying the equation:

400 = -20s.

Dividing both sides by -20:

s = -400/-20,

s = 20 meters.

Therefore, the maximum height attained by the ball is 20 meters.

(c) How long will it take the ball to hit the ground?

To find the time it takes for the ball to hit the ground, we can use the following equation:

s = ut + (1/2)at²,

where:

s = displacement (60 meters downwards),

u = initial velocity (20 m/s upwards),

a = acceleration (acceleration due to gravity, -10 m/s²),

t = time.

Rearranging the equation, we get:

-60 = 20t + (1/2)(-10)t².

Simplifying the equation:

-60 = 20t - 5t².

Rearranging to form a quadratic equation:

5t² - 20t - 60 = 0.

Dividing both sides by 5:

t² - 4t - 12 = 0.

Solving the equation using the quadratic formula, we get:

t = (4 ± sqrt(16 + 4 x 12)) / 2

t = (4 ± 4sqrt(2)) / 2

t = 2 ± 2sqrt(2)

Since time cannot be in negative terms, we ignore the negative value of t.  Therefore, the time it takes for the ball to hit the ground is:

t = 2 + 2sqrt(2) seconds

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[tex] \\[/tex]

(a) How long will it take the ball to pass the man moving in the downwards direction ?

Using Equation of Motion:-

[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{v = u + at}}}}}[/tex]

where:-

Plugging in Values:-

[tex] \large \sf \longrightarrow \: v = u + at[/tex]

[tex] \large \sf \longrightarrow \: 0 = 20 + ( - 10)t \: [/tex]

[tex] \large \sf \longrightarrow \: 0 - 20=( - 10)t \: [/tex]

[tex] \large \sf \longrightarrow \: - 10t = - 20\: [/tex]

[tex] \large \sf \longrightarrow \: t = \frac{ - 20}{ - 10} \\ [/tex]

[tex] \large \sf \longrightarrow \: t = 2 \: secs \\ [/tex]

Therefore, it will take 2 seconds for the ball to pass the man moving in the downward direction.

________________________________________

[tex] \\[/tex]

(b) What is the maximum height attained by the ball ?

→ To solve the given problem, we can use the equations of motion

Using Equation of Motion:-

[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ {v}^{2} = {u}^{2} + 2as}}}}}[/tex]

where:

Plugging in Values:-

[tex] \large \sf \longrightarrow \: {v}^{2} = {u}^{2} + 2as[/tex]

[tex] \large \sf \longrightarrow \: {(0)}^{2} = {(20)}^{2} + 2( - 10)s[/tex]

[tex] \large \sf \longrightarrow \: 0 = 400 + 2( - 10)s[/tex]

[tex] \large \sf \longrightarrow \: 400 + (- 20)s = 0[/tex]

[tex] \large \sf \longrightarrow \: 400 - 20 \: s = 0[/tex]

[tex] \large \sf \longrightarrow \: - 20 \: s = - 400[/tex]

[tex] \large \sf \longrightarrow \: \: s = \frac{ - 400}{ - 20} [/tex]

[tex] \large \sf \longrightarrow \: \: s = 20 \: metres[/tex]

Therefore, the maximum height attained by the ball is 20 meters.

________________________________________

[tex] \\[/tex]

(c) How long will it take the ball to hit the ground ?

Using Equation of Motion:-

[tex] \large\bigstar \: \: \: { \underline { \overline{ \boxed{ \frak{ s= ut + \frac{1}{2}a{t}^{2}}}}}}[/tex]

where:

Plugging in Values:-

[tex] \large \sf \longrightarrow \: s= ut + \frac{1}{2}a{t}^{2}[/tex]

[tex] \large \sf \longrightarrow \: -60= 20t + \frac{1}{2}(-10){t}^{2}[/tex]

[tex] \large \sf \longrightarrow \: -60= 20t + \frac{-10}{2}\times{t}^{2}[/tex]

[tex] \large \sf \longrightarrow \: -60= 20t + (-5)\times{t}^{2}[/tex]

[tex] \large \sf \longrightarrow \: -60= 20t-5\times{t}^{2}[/tex]

[tex] \large \sf \longrightarrow \: 20t-5{t}^{2}+60[/tex]

[tex] \large \sf \longrightarrow \: {t}^{2}-4t-12[/tex]

[tex] \large \sf \longrightarrow \: t=\frac{-b \pm\sqrt{{b}^{2}-4ac}}{2a} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{-(-4) \pm\sqrt{{(-4)}^{2}-4(1)(-12)}}{2(1)} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-4(-12)}}{2} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16-(-48)}}{2} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{16+48}}{2} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{4 \pm\sqrt{64}}{2} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{4 \pm 8}{2} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{4 + 8}{2} \qquad or \qquad t=\frac{4 - 8}{2} [/tex]

[tex] \large \sf \longrightarrow \: t=\frac{12}{2} \qquad or \qquad t=\frac{-4}{2} [/tex]

[tex] \large \sf \longrightarrow \: t=6 \qquad or \qquad t= -2 [/tex]

Since time cannot be negative , Therefore, it will take 6 secs to hit the ground!!

________________________________________

[tex] \\[/tex]

✅

(a)i) When x polarized light is incident on the device, the transmitted light polarization state is given by T

=Jx

= [1 0; 0 1/3] [1; 0]

= [1; 0].ii) When y polarized light is incident on the device, the transmitted light polarization state is given by T

=Jy

= [1/3 0; 0 1] [0; 1]

= [0; 1]

=0,

= 0; Therefore, the eigenvalues of J are λ₁

=1 and λ₂

=1/3. Corresponding to these eigenvalues, we find the eigenvectors by solving (J-λ₁I) p₁

=0 and (J-λ₂I) p₂

=0. Thus, we get: p₁

= [1; 0] and p₂

=[0; 1]. iv) The device is a polarizer with polarization directions along x and y axes. T

= |T|²Io

= 1/3 Io. The reflected beam is also unpolarized, so its intensity is also 1/3 Io.

= 2/3 Io.

=λ/4, E z has maximum amplitude and is in phase with Ey, while at z

=3λ/4, Ez has minimum amplitude and is out of phase with Ey.

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The focal length of 1.50 D reading glasses found on the rack in a pharmacy is 0.67 meters.

The focal length of a lens is a measure of its ability to converge or diverge light. It is commonly denoted by the symbol 'f'. In this case, we are given that the reading glasses have a power of 1.50 D. The power of a lens is the reciprocal of its focal length, so we can use the formula f = 1 / power to determine the focal length.

Substituting the given power of 1.50 D into the formula, we have f = 1 / 1.50. Simplifying this expression, we find that the focal length of the reading glasses is approximately 0.67 meters.

Therefore, the focal length of the 1.50 D reading glasses found on the rack in the pharmacy is 0.67 meters.

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The moment of inertia of the molecule is I = hc / (ΔE * λ). The distance from the atom with mass m to the center of mass of the diatomic molecule can be found using the concept of reduced mass. The reduced mass (μ) takes into account the relative masses of the two atoms in the molecule.

The reduced mass (μ) is given by the formula:

μ = [tex](m_1 * m_2) / (m_1 + m_2)[/tex]

where m1 is the mass of the first atom (m) and m2 is the mass of the second atom (M).

The distance from the atom with mass m to the center of mass (d) can be calculated using the formula:

d =[tex](m_2 / (m_1 + m_2)) * r[/tex]

where r is the distance between the two atoms.

Now, let's consider the energy difference between rotational levels with angular momentum quantum numbers l and (l - 1), where l represents the angular momentum quantum number. The energy difference is given by:

ΔE = ([tex]h^2 / (8\pi ^2))[/tex] * (l / I)

where h is Planck's constant and I is the moment of inertia of the molecule.

To show that the energy difference between rotational levels with quantum numbers l and (l - 1) is[tex]lh^2 / (8\pi ^2I),[/tex]we can substitute (l - 1) for l in the formula and observe the result:

ΔE =[tex](h^2 / (8\pi ^2))[/tex]* ((l - 1) / I)

Simplifying:

ΔE =[tex](h^2 / (8\pi ^2)) * (l / I) - (h^2 / (8\pi ^2I))[/tex]

We can see that this expression matches the formula given in the question, showing that the energy difference between rotational levels with angular momentum quantum numbers l and (l - 1) is lh^2 / (8π^2I).

For the transition from l = 1 to l = 0 in the rotational energy state, the wavelength of the emitted photon (λ) is given as 1.0 × 10^3 m. We can use the equation:

ΔE = hc / λ

where h is Planck's constant and c is the speed of light. Rearranging the equation to solve for I, the moment of inertia of the molecule:

I = hc / (ΔE * λ)

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The focal length of the lens is 6.024 mm. The angular magnification of the telescope is 7.00. The distance L from the ornament that Gwen is located is 3.62 cm.

1. The focal length of the lens in centimeters. The angular magnification M is given by:M = 1 + (25/f)Where f is the focal length of the lens in centimeters. The angular magnification is given as 5.15. Hence,5.15 = 1 + (25/f)f = 25/4.15f = 6.024 mm

2. The angular magnification of the telescope.The formula for the angular magnification of the telescope is given as:M = - fo/feWhere fo is the focal length of the objective lens and fe is the focal length of the eyepiece. The angular magnification is the absolute value of M.M = | - 94.5/13.5 |M = 7.00. The angular magnification of the telescope when you look at the Moon is 7.00.

3. The distance Gwen is located from the ornamentThe distance of Gwen from the ornament is given by the formula:L = (R^2 - h^2)^(1/2) - dWhere R is the radius of the spherical ornament, h is the distance between the center of the ornament and the location of Gwen's image, and d is the distance of Gwen's eye to the ornament. The values of these quantities are:R = 7.9/2 = 3.95 cmh = 1.5 cm (given)d = L (unknown)L = (R^2 - h^2)^(1/2) - dL = (3.95^2 - 1.5^2)^(1/2) - 0L = 3.62 cm (rounded to two decimal places)Hence, the distance L from the ornament that Gwen is located is 3.62 cm.

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I) The phase difference between a ray that arrives at the screen 0.8 cm from the central maximum and a ray that arrives at the central maximum = 8.25 radian.

II) The intensity of the light relative to the intensity of the central maximum at the point on the screen = 1.22 × 10^-3

III)  The order of the bright fringe nearest the point on the screen is 3.

wavelength, λ = 460 nm

Spacing between the slits, d = 0.2 mm

Distance from the slits to a screen, L = 1.2 m

I) The distance of the screen from the central maximum is given by:

x = L λ / d

where, L is the distance from the slits to the screen,

λ is the wavelength of light, and

d is the distance between the slits.

Substituting the given values:

x = (1.2 × 10^3) × (460 × 10^-9) / (0.2 × 10^-3) = 0.276 m

Phase difference, Δϕ = 2πx / λ = 2π(0.276) / (460 × 10^-9) = 8.25 radian

II) The intensity of the light at a point on the screen due to the interference of two waves is given by the formula:

I = 4I_0 cos^2 (Δϕ / 2)

Where, I_0 is the intensity of the light at the central maximum,

Δϕ is the phase difference between two waves.

So, I = 4I_0 cos^2 (Δϕ / 2) = 4 × 1 cos^2 (8.25 / 2) = 1.22 × 10^-3

III) The position of the nth bright fringe is given by:

y_n = nλL / d = (n × 460 × 10^-9 × 1.2) / (0.2 × 10^-3) = 2.76 × 10^-3n m

When y_n = 8 mm = 8 × 10^-3 m, we get the position of the bright fringe nearest the point on the screen.

So, n = (8 × 10^-3) / (2.76 × 10^-3) = 2.9≈3

∴ The order of the bright fringe nearest the point on the screen is 3.

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We know that the coefficient of linear expansion of aluminum, α = 23.1 x 10-6 K-1 Hence,∆L = αL∆T= 23.1 × 10-6 × 688.78 × (253.3 − 131.6)= 4.655 mmThus, the change in length of the wire due to the temperature change is 4.655 mm (rounded to 3 decimal places with u

The length change of an aluminum wire with a diameter of 41.4 mm and 688.78 mm length from a temperature change from 131.6 C to 253.3 C is 4.655 mm. The formula that is used to calculate the change in length of the wire due to temperature change is:∆L

= αL∆T

where, ∆L is the change in length L is the original length of the wireα is the coefficient of linear expansion of the material of the wire∆T is the change in temperature From the provided data, we know the following:Length of the aluminum wire

= 688.78 mm Diameter of the aluminum wire

= 41.4 mm Radius of the aluminum wire

= Diameter/2

= 41.4/2

= 20.7 mm Initial temperature of the aluminum wire

= 131.6 C Final temperature of the aluminum wire

= 253.3 C

We first need to find the coefficient of linear expansion of aluminum. From the formula,α

= ∆L/L∆T We know that the change in length, ∆L

= ?L = 688.78 mm (given)We know that the initial temperature, T1

= 131.6 C

We know that the final temperature, T2

= 253.3 C.We know that the coefficient of linear expansion of aluminum, α

= 23.1 x 10-6 K-1 Hence,∆L

= αL∆T

= 23.1 × 10-6 × 688.78 × (253.3 − 131.6)

= 4.655 mm Thus, the change in length of the wire due to the temperature change is 4.655 mm (rounded to 3 decimal places with units).

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