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Aleksandr-060686 [28]

3 years ago

6

8. A solid wooden door, 90 cm wide by 2.0 m tall, has a mass of 35 kg. It is open and at rest. A small 500-g ball is thrown perp

endicular to the door with a speed of 20 m/s and hits the door 60 cm from the hinged side, causing it to begin turning. The ball rebounds along the same line with a speed of 16.0 m/s relative to the ground. If the momentum of inertia of the door around the hinge is I=1/3 Ma 2 , where a is the width of the door, how much energy was lost during this collision?

1 answer:

insens350 [35]3 years ago

7 0

Answer:

Explanation:

Kinetic energy of ball

= .5 x .5 x 20²

= 100 J

Original kinetic energy of door = 0

Total kinetic energy before ball hitting the door

= 100 J

We shall apply law of conservation of momentum to calculate angular velocity of the door after ball hitting it.

change in angular momentum of ball

= mvr - mur , u is initial velocity and v is final velocity of ball

= .5 ( 20 + 16 ) x .06

= 1.08

Change in angular momentum of door

= I x ω - 0

1/3 x 35 x .09² x ω

= .0945 x ω

so

.0945 x ω = 1.08

ω = 11.43

rotational K E of door after collision

= 1/2 I ω²

= .5 x .0945 x 11.43 ²

= 6.17 J

Kinetic energy of ball after collision

= 1/2 x .5 x 16²

= 64

Total KE of door and ball

= 64 + 6.17

= 70.17 J

LOSS OF ENERGY

= 100 - 70.17 J

= 29.83 J

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Answer:

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Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

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