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pantera1 [17]
2 years ago
6

A railroad car of mass 2.50∙10^4 kg is moving at a speed of 4.00 m/s. It collides and couples with three other coupled railroad

cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.00 m/s.
(a)What is the speed of the four cars after the collision?
(b)How much kinetic energy is lost in the collision? Where does this energy go?
Physics
1 answer:
Vinvika [58]2 years ago
8 0

Answer:

a) v = 1.00 m/s

b) ΔKE = 150 kJ

Lost  energy makes heat, noise, vibration in the system

Explanation:

conservation of momentum

let m be the mass of one car

m(4.00) + 3m(0) = 4mv

v = 1.00 m/s

ΔKE = ½m(4.00)² + ½(3m)(0.00)² - ½(4m)(1.00)²

ΔKE = 8m + 0m - 2m = 6m J

ΔKE = 6(2.50e4) = 150 kJ

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"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two firs
Fantom [35]

Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

6 0
3 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0
2 years ago
If one of two interacting charges is doubled, the force between the charges will _____________.
malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

F=\frac{k q1 q2}{r^{2}}

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get F'=\frac{k (2q1) q2}{r^{2}}

F'= 2 F

Thus the force gets doubled.

4 0
3 years ago
7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca
Misha Larkins [42]

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

8 0
2 years ago
Which type of force if due to the masses of objects.
statuscvo [17]

Answer: C

Explanation: weak nuclear

7 0
2 years ago
Read 2 more answers
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