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algol [13]
2 years ago
5

We dropped a golf ball from 100 cm in class. it bounced back to 60cm. why did it not return to the 100 cm mark?

Physics
1 answer:
Elena-2011 [213]2 years ago
6 0

Because the golf ball has (0 < e < 1) of elasticity therefore it did not experience the perfect elastic collision, the golf ball experienced an inelastic collision.

An elastic collision is a collision where there is no energy loss. It can be described by elasticity value. When an object has an elasticity value of 1, the speed of the object before the collision or after the collision is the same. The elasticity can be determined as

e = KEafter / KEbefore

where e is the elasticity of the object

when the object has elasticity 0 < e < 1, the object will partially lose the energy at collision.

For more on elasticity at: brainly.com/question/17879061

#SPJ4

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A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from
kaheart [24]

Answer:

zero

Explanation:

In this system, only conservative forces act. Therefore, the mechanical energy, that is, the sum of the kinetic energy and the potential energy, remains constant. When the mass is at its maximum displacement from equilibrium, its potential energy is maximum, therefore, its kinetic energy is minimal, that is to say, that its instantaneous velocity at that point is zero.

6 0
4 years ago
A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is
Tasya [4]

Answer:

C) T

Explanation:

M = Mass attached to an ideal spring

A = Amplitude of the motion

T = Time period of oscillation

k = Spring constant of the spring

A = Amplitude of the motion

Time period of oscillation of the mass attached to the spring is given as

T = 2\pi \sqrt{\frac{M}{k} }

So we see that the time period does not depend on the amplitude. hence the period of oscillation remains the same.

8 0
4 years ago
A gymnast jumps directly up onto a beam that is 1.5 m high. From the beam, she jumps 0.5 m straight up and lands back on the bea
tatiyna

Answer:

D. 2.5 m

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1.5 m up to the beam + 0.5 m up + 0.5 m down = 2.5 m total

5 0
4 years ago
1. A bicycle initially moving with a velocity
ki77a [65]

Answer:

\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}

Explanation:

We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

v_f= v_i+ at

In this formula, v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time.

The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

\bullet \  v_i = 5.0 \ m/s \\\bullet \  a= 2\ m/s^2\\\bullet \  t= 5  \ s

Substitute the values into the formula.

v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)

Solve inside the parentheses.

  • \frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s

v_f= 5.0 \ m/s + (10 \ m/s)

Add.

v_f= 15 \ m/s

The units can also be written as:

v_f= 15 \ m*s^{-1}

The bicycle's final velocity is 15 meters per second.

7 0
3 years ago
Which describes the motion of the media for a surface wave?
Sidana [21]

Answer:

a. the motion is both parallel and perpendicular

Explanation:

I did grad point :)

4 0
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