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 perpendicular 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?
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Ответ:
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
Ответ:
to trap pathogens and to prevent pathogens from moving into deep tissue!
Explanation:
I think please tell me if I am wrong