How do I solve Q.14 & Q.15

To make the next delivery, Sam-I-Am goes 45 mi/h for 2 hours. How far did he travel in that time?

andrezito [222]

Hello =D

This problem is about cinematic

So

V = 45 mi/h

t = 2 h

Then

V= X/t

X = V*t

Then

X = (45)*(2)

X = 90 mi

Best regards

5 0

3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?

Crazy boy [7]

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

6 0

3 years ago

A radio wave has a speed of 3.00 x10^8 and a frequency of 100 MHz. What is the wavelength?

dlinn [17]

Answer:

Wavelength of radio is wave is 3 m

Explanation:

Wavelength of radio is wave is

$\lambda=\frac{v}{f}$

where

$v=3 \times 10^8 m\s\\f=100 Mhz$

wavelength is

$\lambda=\frac{3\times 10^8}{100\times 10^6} \\\\\lambda=3 m\\$

5 0

3 years ago

If a 15g irregular shaped object is submerged in a graduated cylinder of water, the level rises from 10 ml to 25 ml, what is the

SashulF [63]

Answer:

one im so sry i have no idea. I have been researshing for about 30min and i cant find anything im so sry:/

Explanation:

6 0

3 years ago