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Vanyuwa [196]
3 years ago
13

What is the sl unit for momentum

Physics
1 answer:
mario62 [17]3 years ago
8 0
Momentum is mass times velocity and mass is measured in kg and velocity is measured per second so it really depends.
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D, electron, the nucleus is not a single particle to begin with, the proton has a positive charge, a neutron has a neutral charge or no charge, and an electron has a negative charge
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How often do the earth's magnetic poles switch?
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Reversals are the rule, not the exception. Earth has settled in the last 20 million years into a pattern of a pole<span> reversal about every 200,000 to 300,000 years, although it has been more than twice that </span>long<span> since the last reversal.</span>
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Chanice drives her scooter 7 kilometers north. she stops for lunch and then drives 5 kilometers and then 1 km east again. what d
lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

d = d_1 + d_2 + d_3

d = 7 km + 5 km + 1 km

d = 13 km

Now in order to find the displacement we will show all with their directions

\vec d_1 = 7 + 5 = 12 km towards North

\vec d_2 = 1 km towards East

So total displacement is

\vec d = \vec d_1 + \vec d_2

\vec d = 12 \hat j + 1 \hat i

so net displacement will be

d = \sqrt{12^2 + 1^2} = 12.04 km

so displacement is 12.04 km

6 0
3 years ago
Air in the atmosphere is heated by the ground this warm air then rises and cooler air falls this is an example of what type of p
Alona [7]
The answer for both is ‘B’
8 0
3 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
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