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ZanzabumX [31]
3 years ago
9

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an

d make it vibrate in the fundamental frequency once again. The rubber band is made so that doubling its length doubles the tension and reduces the mass per unit length by a factor of 2. The new frequency will be related to the old by a factor of:
Physics
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = \sqrt{\frac{T}{\frac{M}{L}}}

Frequency (F₁) = \frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = \frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

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Darya [45]

Answer:

Explanation:

If Ek is the kinetic energy and m is the mass and v is the velocity then v can be calculated as follows

Ek= 1/2 ×( m × v² )

2Ek= mv²

2Ek/m = v²

v =√(2Ek/m)

m = 0.1 kg

v= √(2x8/0.1)= 12.65 m/s

5 0
3 years ago
A 1.0 kg ball falls from rest a distance of 19.6 m.
Nimfa-mama [501]

Answer:

192.08J

19.6m/s

Explanation:

Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.

PE=mgh

=(1)(9.8)(19.6)

=192.08J

v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.

v²=u²+2as

v²=0²+2(9.8)(19.6)

v=√384.16

=19.6m/s

6 0
2 years ago
Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (
Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}

9.28 + 8.3 x = 0

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5 0
3 years ago
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castortr0y [4]

Answer:

the source of sound moves towards an observe

Explanation:

The Doppler effect is related to waves such as sound or light. the effect causes an increase or decrease in the frequency of sound light or other waves when the souces either move towards or away from the observer. For example the siren of the train to a person on the platform, the redshift seen by astronomers.

Therefore, The Doppler shift can be observed when the source of sound moves towards an observer From a place closer to the observer than the last wave's crest, each consecutive wave crest is sent. Each wave therefore, takes a little less time than the preceding wave to reach the observer.

6 0
2 years ago
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Keith_Richards [23]

Answer:

1. Recollapsing universe

2. Critical universe

3. Coasting universe

Explanation:

Recollapsing universe has dark matter density greater than critical density. While critical universe has its matter density equal to the critical sensity. Coasting universe on the other hand has much smaller matter density compared to critical density.

Note that the critical density is approximately 10^-20 grams/cm3

3 0
3 years ago
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