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3 years ago

7

A 40kg kid riding their bike down the street at 5m/s reaches the edge of a hill and coast down to the bottom. If the hill was 10

m high, how much mechanical energy does the rider have at the bottom of the hill?

1 answer:

Romashka [77]3 years ago

6 0

Answer

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Explanation:v

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Why do you think it is important that the U.S. Constitution defines citizenship?

velikii [3]

Answer:

It's important that the U.S. Constitution defines citizenship because it guarantees it to people who are born within the United States.

Explanation:

It"subjects to the jurisdiction thereof".

5 0

3 years ago

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

3 years ago

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average forc

Natali5045456 [20]

Answer:

<h2>9.3kN</h2>

Explanation:

Step one:

given data

mass of bullet= 0.02kg

speed v=700m/s

time taken =1.5ms= 0.0015 seconds

Step two:

we know that from the first law

F=ma-----1 first law of motion

also, we know that

a=v/t----2

put a=v/t in equation 1 we have

F=mv/t

Step three:

substitute our given data to find force

F=0.02*700/0.0015

F=14/0.0015

F=9333.33N

F=9.3kN

<u>The average force exerted is 9.3kN</u>

7 0

2 years ago

What is the magnetic force on a proton that is moving at 4.5 107 m/s to the right through a magnetic field that is 1.6 T and poi

pickupchik [31]

The answer is letter C. 1.2 10^-11 N up
Solution:
F= Bqvsin(theta)
theta = sin 90 = 1
F= 1.4 T * 1.6x10^-19 * 5.2x10^7 ms^-1
F= 1.16 x 10^-11 N
Then the direction is upward.

6 0

3 years ago

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True or false? Most forces occur in pairs, such as when

statuscvo [17]

This is TRUE everything has a action the force from your feet pushes and the floor pushes back

8 0

3 years ago