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Zarrin [17]
3 years ago
10

Two cars go through 2 different crashes. Car 1 experiences a 500 Ns impulse for a duration of 15s, while car 2 experiences that

same Impulse for a duration of 5s.
a) Which car experiences the most force during their collision? (Show your work for both cars and round your answers to 2 decimal points) (include what equation you are using) 5pts

b) Why is the number in part a important to know? Use your own understanding and the explanation from class. (Use real world examples) 5pts

Write your answer:
Physics
1 answer:
tatuchka [14]3 years ago
5 0
B is your answer hope this helps;)
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The magnetic field and force is weakest in the center of the magnet. The correct answer is C. 
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Give the number of significant figures: 832.000 km
Galina-37 [17]
It has three significant figure
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A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en
Maksim231197 [3]

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient) 

W= f x d = uNd ; N=mg 
Delta E = 1/2 mV^2 -1/2mVi^2 

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term) 

This will then give us: 

1/2Vi^2-ugd = 1/2V^2 

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s 

 

6 0
4 years ago
What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?
statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000  x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J

Kinetic energy is given as;

K.E = ¹/₂mv²

where;

v is speed of the electron

K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} =  (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} =  (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

6 0
3 years ago
1. A stationary 50 tonne submarine fires a missile of mass 40kg at
White raven [17]

The velocity of the submarine immediately after firing the missile is 0.0104 m/s

Explanation:

Mass of the submarine M=50 tonne=50\times 1000=50000kg

Mass of the missile m=40  kg

velocity of the missile v= 13m/s

we have to calculate the velocity of the submarine after firing

This is the recoil velocity and its expression is derived from the law of conservation of momentum

recoil velocity of the submarine

V=-mv/M\\=-40\times 13/50000\\=0.0104 m/s

8 0
3 years ago
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