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steposvetlana [31]
3 years ago
6

A juggler throws a ball straight up into the air. The ball remains in the air for a time (t) before it lands back in the juggler

s hand. What is the acceleration of the ball during the entire time the ball is in the air?
Physics
1 answer:
natka813 [3]3 years ago
5 0

Answer:

9.8 m/s^2, downward

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity (downward)

According to Newton's second law,

F=ma

where F is the net force on the object and a is its acceleration. Rearranging for a,

a=\frac{F}{m}

As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:

a=\frac{mg}{m}=g

Therefore, the ball has a constant acceleration of 9.8 m/s^2 downward for the entire motion.

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For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m
Leviafan [203]

Answer:

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

\Sigma F = F-R = 0 (Eq. 1)

Where:

F - Forward trust, measured in newtons.

R - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: (F = 6.85\times 10^{5}\,N)

R = F

R = 6.85\times 10^{5}\,N

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

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