Question 3 Please, i need help! Thank you

What is true for ALL of the examples of electromagnetic waves?

nasty-shy [4]

A they all move the same speed in a vacuum this is because in a the photons don't bump in to anything

5 0

3 years ago

Yurem is pulling a wagon across the playground with a force of 10 N. He asks Elianna to help. She agrees and pushes the back of

salantis [7]

Answer:

22 N applied force

Explanation: Since they are both pushing the wagon in the same direction the force adds up.

7 0

3 years ago

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A mover pushes a 30 kg crate across a level florr with a force of 250 N, but the crate accelerates at a rate of only 5.83 m/s sq

arlik [135]

Force of friction=75 N

Explanation:

we use Newton's second law of motion

F- Ff= ma

F = applied force=250 N

Ff= force of friction

a= acceleration= 5.83 m/s²

m= mass=30 kg

250- Ff= 30 (5.83)

Ff=250-30(5.83)

Ff=250-175

Ff=75 N

7 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

4 years ago

A 20 gram bullet, traveling horizontally with a velocity of magnitude 350 m/s, is fired into a wooden block with a mass of 0.75,

sertanlavr [38]

Answer:

The kinetic energy is 10.65 joules.

Explanation:

This case can be considered as a collision if we neglect the frictional forces the ball experiences traveling through the wooden block because are lower than the impact forces, then we can use conservation of linear momentum to model this situation considering the system block-bullet.

Linear momentum states total initial momentum (pi) is equal to the final momentum (pf), this is:

$p_f=p_i$

Total momentum is the sum of individual momentums, in our case the momentum of the ball (pb) plus the momentum of the wooden block (pw), then:

$p_{bf}+p_{wf}=p_{bi}+p_{wi}$

Linear momentum of an object is defined as mass times velocity, then:

$m_bv_{bf}+m_wv_{wf}=m_bv_{bi}+m_wv_{wi}$

With vb the velocities of the bullet, mb the mass of the bullet, vw the velocities of the wooden block and mw the mass of the wooden block. Note that the wooden block is initially at rest, that means the initial velocity of the wooden block is zero, then:

$m_bv_{bf}+m_wv_{wf}=m_bv_{bi}$

Solving the equation for the final velocity of the block:

$v_{wf}=\frac{m_bv_{bi} - m_bv_{bf}}{m_w}=\frac{m_b(v_{bi} - v_{bf})}{m_w}$

$v_{wf}=\frac{0.02(350- 150)}{0.75} =5.33 \frac{m}{s}$

(Important: i. The sign of the velocities is important, so we chose positive direction as the direction the bullet moves initially, positive velocities means they are in this same direction.

ii. I assume 0.75 kg for the mass of the wooden block because you didn't provide its unities)

Now with the velocity of the bullet after the collision we can find the kinetic energy using the equation for kinetic energy:

$K=\frac{m_wv_{wf}^2}{2}=\frac{(0.75)(5.33)^2}{2}$

$K=10.65 J$

4 0

3 years ago

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