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andrew-mc [135]
3 years ago
13

A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the

velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward. the velocity is zero, the acceleration is zero, and the force of gravity acting on the ball is zero. the velocity is zero, the acceleration is zero, and the force of gravity acting on the ball is directed downward. the velocity and acceleration reverse direction, but the force of gravity on the ball remains downward. the velocity, acceleration, and the force of gravity on the ball all reverse direction.
Physics
1 answer:
nignag [31]3 years ago
7 0

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

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A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma
Alex

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}

in this case:

3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s

the initital kinetic energy is:

K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ

and the final:

K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ

The energy lost is given by:

E_l=|K_f-K_i|\\E_l=23kJ

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Answer:

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