Recall that if the colliding objects have hard surfaces , the collision is elastic

Find an expression for the electric field e⃗ at the center of the semicircle. hint: a small piece of arc length δs spans a small

ahrayia [7]

Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by

C = 2L = 2*pi*R ---> R = L/pi 

Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. 

we can define a small charge dq as 

dq = l*ds = l*R*da 

So the electric field can be written as: 

dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) 

dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) 

E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) 

E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)

8 0

3 years ago

An ideal monatomic gas at temperature T is held in a container. If the gas is compressed isothermally, that is at constant tempe

OlgaM077 [116]

Answer:

a) 0 J

b) W = nRTln(Vf/Vi)

c) ΔQ = nRTln(Vf/Vi)

d) ΔQ = W

Explanation:

a) To find the change in the internal energy you use the 1st law of thermodynamics:

$\Delta U=\Delta Q-W$

Q: heat transfer

W: work done by the gas

The gas is compressed isothermally, then, there is no change in the internal energy and you have

ΔU = 0 J

b) The work is done by the gas, not over the gas.

The work is given by the following formula:

$\\W=nRTln(\frac{V_f}{V_i})$

n: moles

R: ideal gas constant

T: constant temperature

Vf: final volume

Vi: initial volume

Vf < Vi, then W < 0 and the work is done on the gas

c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.

The amount of heat is equal to the work done W

d)

$\Delta U = \Delta Q-W\\\\0=\Delta Q-W\\\\\Delta Q=W=nRTln(\frac{V_f}{V_i})$

5 0

3 years ago

You drive a car 1600 ft to the east, then 2500 ft to the north. The trip took 2.5 minutes. What was the magnitude of your averag

elena-14-01-66 [18.8K]

Answer:

$Velocity=6.03m/s$

Explanation:

Given data

Time t=2.5 minutes=150 seconds

Distance A=1600 ft=487.68 m........east

Distance B=2500 ft=762m ........north

To find

Average velocity

Solution

First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse

So

$A^{2}+B^{2}=C^{2}\\ C=\sqrt{A^{2}+B^{2}}\\ C=\sqrt{(487.68m)^{2}+(762m)^{2}}\\ C=904.7m$

$Velocity=\frac{Distance}{Time}\\Velocity=\frac{904.7m}{150s}\\Velocity=6.03m/s$

7 0

2 years ago

A stone is thrown horizontally from the top of an inclined plane (angle of inclination θ). How would I find the initial speed of

Katen [24]

Answer:

S = V t where S is the horizontal distance traveled

1/2 g t^2 = H where H is the vertical distance traveled

t^2 = 2 H / g

V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations

tan theta = H / S

V^2 = S g / (2 tan theta)

Using S = L cos theta

V^2 = L g cos theta / (2 tan theta)

Giving V in terms of L and theta

7 0

2 years ago

A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring project

Gwar [14]

Answer:

$x' = 10 x$

Explanation:

By energy conservation we know that spring energy is converted into kinetic energy of the block

so we will have

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

so we will have

$v = \sqrt{\frac{k}{m}}(x)$

now we will have same thing for another mass 4m which moves out with speed 5v

so we have

$5v = \sqrt{\frac{k}{4m}}(x')$

now from above two equations we have

$\frac{5v}{v} = \frac{x'}{2x}$

so we have

$x' = 10 x$

3 0

2 years ago