PHYSICS COLLEGE

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that when placed in the field the sphere is in a static situation (all the forces on the sphere cancel). If the thread is horizontal, find the magnitude and direction of the electric field. The sphere has a mass of 0.018 kg and contains a charge of + 6.80 x 103 C. The tension in the thread is 6.57 x 10-2 N. Show your work and/or explain your reasoning. (20 pts)

Answers

Answer 1

Answer:

Answer:

$E = 9.66* 10^(-6) N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 * 10^(-2) N$

$Q = 6.80 * 10^3 C$

now we have

$6.57 * 10^(-2) = (6.80 * 10^3)E$

$E = 9.66* 10^(-6) N/C$

direction is Horizontal

Answer 2

Answer:

Final answer:

The magnitude of the electric field on the charged sphere in this scenario is approximately 1.17 x 10^-5 N/C. The direction of the electric field is horizontal, which is the same direction as the tension in the thread.

Explanation:

To start, we can use the equilibrium condition where the tension in the thread is equal to the force due to the electric field and gravity on the sphere. The formula to calculate the electric force is F = qE, and the gravitational force is F = mg, where F is the force, q is the electric charge, E is the electric field, m is the mass of the object, and g is the gravity constant.

Tension - electric force - gravitational force equals zero: T - F_electric - F_gravity = 0. We fill in the previous formulas: T - qE -mg = 0. This can be rearranged to E = (T + mg) / q.

In this case, the sphere's mass m is 0.018 kg, the tension T is 6.57 x 10^-2 N, and the sphere's charge q is 6.80 x 10^3 C, and we use g = 9.81 m/s². So, E = ((6.57 x 10^-2) + (0.018 * 9.81)) / 6.80 x 10^3.

This leads to an electric field magnitude of approximately 1.17 x 10^-5 N/C. The direction of the electric field is the same as the direction of the tension, which is horizontal due to the thread being horizontal.

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An isotropic point source emits light at wavelength 500 nm, at the rate of 185 W. A light detector is positioned 380 m from the source. What is the maximum rate ∂B/∂t at which the magnetic component of the light changes with time at the detector's location?

Answers

Answer:

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

Explanation:

Given that

An isotropic point source emits light at a wavelength $\lambda$ = 500 nm

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity of the wave , which is = $(Power )/(Area)$

= $(Power)/(4 \pi r ^2)$

= $(185 \ W)/( 4 \pi (380)^2)$

= $1.0195*10^(-4) \ W/m^2$

Now;

The amplitude of the magnetic field is calculated afterwards by using poynting vector

i.e

$I = ((c)/(2 \mu_0 ))B_(max^2)$

$B_(max^2) = ((2 \mu_0 I)/( c))$

$B_(max^2) = ((2 *4 \pi *10^(-7)*1.0195*10^(-4))/( 3*10^8))$

$B_(max^2) = 8.5409*10^(-19)$

$B_(max) = \sqrt {8.5409*10^(-19)}$

$B_(max) = 9.242*10^(-10)$

The magnetic field wave equation can now be expressed as;

$B = B_(max) sin (kx - \omega t)$

Taking the differentiation

$(dB)/(dt)= - \omega B_(max) \ cos ( kx - \omega t)$

The maximum value ;

$(dB)/(dt) = \omega B _(max)$

where ;

$\omega = 2 \pi f\n\omega = (2 \pi c)/(\lambda)$

then

$(dB)/(dt) = (2 \pi c)/(\lambda) B _(max)$

$(dB)/(dt) = (2 \pi 3*10^8*9.242*10^(-10))/(500*10^(-9))$

$(dB)/(dt) = 3484751.917$

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

The maximum rate $(∂B/∂t)$ at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

To find the maximum rate at which the magnetic component of the light changes with time at the detector's location, you can use the formula for the rate of change of magnetic field due to an electromagnetic wave. The formula is given by:

$∂B/∂t = (2π / λ) * E * c$

Where:

$∂B/∂t$ is the rate of change of the magnetic field.

λ is the wavelength of the light.

E is the electric field strength.

c is the speed of light in a vacuum, approximately $3 x 10^8 m/s.$

You have the wavelength (λ) as 500 nm, which is 500 x 10^-9 meters, and the electric field strength (E) can be calculated using the power (P) and the distance (r) from the source. The power emitted by the source is 185 W, and the distance from the source to the detector is 380 m.

First, calculate the electric field strength (E):

$E = sqrt(P / (2π * r^2))$

$E = sqrt(185 W / (2π * (380 m)^2))$

$E = sqrt(185 W / (2π * 144400 m^2))$

$E ≈ 6.325 x 10^-5 N/C$

Now, you can calculate the rate of change of the magnetic field:

$∂B/∂t = (2π / λ) * E * c$

$∂B/∂t = (2π / (500 x 10^-9 m)) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ (3.77 x 10^15 Hz) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ 6.8 x 10^9 T/s$

So, the maximum rate at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

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Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty. The weight of the elevator cabin is partially balanced by a 400-kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the elevator well. Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless, determine (a) the power required while the fully loaded cabin is rising at a constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. What would your answer be to (a) if no counterweight were used? What would your answer be to (b) if a friction force of 800 N has developed between the cabin and the guide rails?

Answers

Answer:

Part a)

$P = 4.71 * 10^3 Watt$

Part b)

$P = 2.94 * 10^3 W$

Part c)

$P = 9.4 * 10^3 W$

Part d)

$P = 3.9 * 10^3 W$

Explanation:

Part a)

When cabin is fully loaded and it is carried upwards at constant speed

then we will have

net tension force in the rope = mg

$T = (800)(9.81)$

$T = 7848 N$

now it is partially counterbalanced by 400 kg weight

so net extra force required

$F = 7848 - (400 * 9.81)$

$F = 3924 N$

now power required is given as

$P = Fv$

$P = 3924 (1.2)$

$P = 4.71 * 10^3 Watt$

Part b)

When empty cabin is descending down with constant speed

so in that case the force balance is given as

$F + (150 * 9.8) = (400 * 9.8)$

$F = 2450 N$

now power required is

$P = F.v$

$P = (2450)(1.2)$

$P = 2.94 * 10^3 W$

Part c)

If no counter weight is used here then for part a)

$F = 7848 N$

now power required is

$P = F.v$

$P = 7848 (1.2)$

$P = 9.4 * 10^3 W$

Part d)

Now in part b) if friction force of 800 N act in opposite direction

then we have

$F + (150 * 9.8) = 800 +(400 * 9.8)$

$F = 3250 N$

now power is

$P = (3250)(1.2)$

$P = 3.9 * 10^3 W$

Convert 56km/h to m/s.

Answers

Explanation:

15.556 metres per second

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by:

Vx,f = sqrt (2*g*x*sin(Q))

What best describes a societal law

Answers

Answer:

Societal laws are based on the behavior and conduct made by society or government.

hope it helps.

stay safe healthy and happy.

A relaxed biceps muscle requires a force of 25.4 N for an elongation of 3.20 cm; under maximum tension, the same muscle requires a force of 520 N for the same elongation. Find Young's modulus (Pa) for the muscle tissue under each of these conditions if the muscle can be modeled as a uniform cylinder with an initial length of 0.200 m and a cross-sectional area of 50.0 cm2.