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3 years ago

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A ball thrown vertically upward reaches a certain height and comes down again. What can you say about its kinetic energy at the

maximum height?
It is independent of the height.
It is at its minimum but is not zero.
It is at its maximum.
It is zero.

2 answers:

MrRissso [65]3 years ago

4 0

The K.E. at the maximum height is zero, no matter what that height is.

stellarik [79]3 years ago

3 0

The kinetic energy at the maximum height is zero

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Goshia [24]

Answer:

Environmental Protection Agency (EPA)

Explanation:

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3 years ago

Dimensional Analysis : 3 days to seconds

LuckyWell [14K]

Hi,

To convert 3 days to seconds write this.

1h = 3600s
24h = 3600 · 24 = 86400
3 days = 3 · 86400 = 259200sec

Hope this helps.
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3 years ago

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A ball with mass of 0.050 kg is dropped from a height of h1 = 1 .5 m. It collides with the floor, then bounces up to a height of

Iteru [2.4K]

Answer:

Explanation:

Impulse of reaction force of floor = change in momentum

Velocity of impact = √ 2gh₁

= √ 2 x 9.8 x 1.5 = 5.4 m /s.

velocity of rebound = √2gh₂

= √ 2x 9.8 x 1

= 4.427 m / s.

Initial momentum = .050 x 5.4 = .27 kg m/s

Final momentum = .05 x 4.427 = .22 kg.m/s

change in momentum = .27 - .22 = .05 kg m/s

Impulse = .05 kg m /s

Impulse = force x time

force = impulse / time

.05 / .015 = 3.33 N.

kinetic energy = 1/2 m v²

Initial kinetic energy = 1/2 x .05 x 5.4²

= 0.729 J

Final Kinetic Energy =1/2 x .05 x 4.427²

= 0.489 J

Change in Kinetic energy =0 .24 J

Lost kinetic energy is due to conversion of energy into sound light etc.

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3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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3 years ago