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jeyben [28]
3 years ago
13

A hockey puck sliding on a frictionless surface strikes a box at rest. After the collision, the two objects stick together and m

ove at some final speed.
Which of the statements describes the change in momentum and energy of the puck during the collision?

a) The puck loses some, but not all, of its original momentum and mechanical energy.
b) One cannot determine if the momentum or energy of the puck is conserved without knowing the final velocities and masses of the two objects.
c) The puck loses some momentum in the collision but conserves its mechanical energy.
d) The puck conserves its original momentum but loses all of its mechanical energy.
e) The puck conserves its original momentum and mechanical energy.
f) The puck conserves its original momentum but loses some, but not all, of its mechanical energy.
Physics
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

Answer:

f) The puck conserves its original momentum but loses some, but not all, of its mechanical energy.

 Explanation:

It is a case of perfectly inelastic collision . So momentum will be conserved because no external force acts on them during the collision . But there will be loss of energy ( kinetic energy ) . It will be in the form of sound or heat that is produced during collision. They will still have some kinetic energy even after the collision.

 

Explanation:

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EastWind [94]

Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

a)

We know that

Electric potential difference  ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

b)

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J

ΔU =18.79 mJ

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Answer:

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3 years ago
a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by
lisov135 [29]

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

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height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

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A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex
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Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

v_i = v_o

now at the end of 80 s let the speed is

v_f = v_1

after another 120 s let the speed will be

v_f' = v_2

now we know that

d = \frac{v_i + v_f}{2} (t)

d = \frac{v_o + v_1}{2}(80)

1000 = 40(v_o + v_1)

also we know that

v_1 - v_o = a(80)

also we have

1000 = \frac{v_1 + v_2}{2}(120)

1000 = 60(v_1 + v_2)

now we can say

(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}

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a = -0.042 m/s^2

Part b)

now we have

v_1 + v_o = 25

v_1 - v_o = (-0.042)(80)

v_1 = 10.83 m/s

so the starting velocity of the trip is

v_o = 25 - 10.83 = 14.17 m/s

now speed after t = 200 s is given as

v_2 = v_o + at

v_2 = 14.17 - (0.042)(200)

v_2 = 5.77 m/s

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