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Rashid [163]
3 years ago
9

When a basketball player dribbles a ball, it falls to the floor and bounces up. Is a force required to make it bounce? Why? If a

force is needed, what is the agent involved?
Physics
1 answer:
Marizza181 [45]3 years ago
5 0
Energy is the one that is stored in the ball when it drops. Just before it hits the ground, the energy depends on the mass of the ball and its velocity. When the ball hits, it is compressed and the energy is stored in the compression of the air in the ball and the elasticity of the material that the ball is made from. Some is also converted to heat. The stored energy in the ball causes a force to make the ball back into a round shape and this force presses against the propels and floor the ball back up. The small amount lost as heat is the reason that the ball bounces up with less energy than when it hit.
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X=(150^2•sin(2•42))/9.8
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Answer:

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Explanation:

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Explanation:

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A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower
charle [14.2K]

Answer:

  She will make the jump.

Explanation:

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

   Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

   77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds

So she will cover 77 m in 2.85 seconds

 Now considering vertical motion, up direction as positive

    Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 m/s^2, time = 2.85

    Substituting

           s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m

  So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

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6 0
3 years ago
Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls
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Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 =  8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

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