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suter [353]
3 years ago
9

You are tossing a ball directly up into the air. Before the ball leaves your hand, you are exerting a force directed against the

force of gravity. If you apply a net force of 10 N to a 1 kg ball, what is the normal force acting on the ball?
Physics
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

F n = 0.2 N

Explanation:

given,                                      

you are exerting force of 10 N on the ball.

mass of the ball = 1 kg              

acceleration due to gravity  = 9.8 m/s²

normal force on the ball = ?          

normal force is force exerted by the object to counteract the force from other object.                

normal force acting on the ball will be

F n = F - mg                          

F n = 10 - 1 × 9.8                        

F n = 10 -9.8                    

F n = 0.2 N            

Hence, normal force acting on the ball is equal to 0.2 N

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The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

s=ut+ \frac{1}{2} at^2

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

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e. Only(a) and (b) above are correct

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3 years ago
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