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Morgarella [4.7K]
3 years ago
8

Why does the velocity change when the boy hits the ball

Physics
2 answers:
MrRa [10]3 years ago
7 0

Answer:

Acceleration from gravity is always constant and downward, but the direction and magnitude of velocity change. At the highest point in its trajectory, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth

Explanation:

juin [17]3 years ago
4 0

Answer:

Freefall is a special case of motion with constant acceleration because acceleration due to gravity is always constant and downward. This is true even when an object is thrown upward or has zero velocity.

When a ball is thrown up in the air, its velocity is initially upward. Since gravity pulls the object toward the earth with a constant acceleration g, the magnitude of velocity decreases as the ball approaches maximum height. At the highest point in its route, the ball has zero velocity, and the magnitude of velocity increases again as the ball falls back toward the earth.

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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
  • For this, we can equate the net horizontal force.

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

4 0
2 years ago
Read 2 more answers
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