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Kaylis [27]
3 years ago
12

A 12 g plastic ball is dropped from a height of 2.5 m and is moving at 3.2 m/s just before it hits the floor. How much mechanica

l energy was lost during the fall?
Physics
1 answer:
Nataly [62]3 years ago
4 0

Answer:

0.24

Explanation:

Mass of ball= 12g=0.012Kg

height of ball= 2.5m

velocity of ball before falling= 3.2m/s

potential energy of the ball=mgh= 0.012*10*2.5=0.3J

kinetic energy of the ball=0.5*mv^{2}=0.5*0.012*3.2*3.2=0.6J

Loss in mechanical energy during the fall= potential energy- Kinetic energy= 0.3-0.06=0.24J

note: During the fall, the potential energy of the ball is converted to kinetic energy. the loss in energy is due to air resistance.

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3 years ago
A box is being pushed from the left with a force of 5 N. No force is being applied from the right.
Sergio039 [100]

Answer:

The box will move to the right with increasing speed

Explanation:

According to Newton's second law:

F=ma (1)

where

F is the net force on an object

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a is its acceleration

For the box in this problem, there is only one force acting on it: a force of 5 N from the left (so, to the right). This means that according to (1) the box will experience an acceleration: the direction of the acceleration is the same as the direction of the net force (so, to the right), therefore the box will move to the right and its speed will continue increasing due to fact that it is being accelerated.

6 0
3 years ago
A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,
NISA [10]

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N      

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity g=9.8m/sec^2

Coefficient of static friction \mu _s=0.584

Coefficient of kinetic friction \mu _k=0.399

(a) Static friction force is given by F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

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3 0
3 years ago
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