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JulijaS [17]
4 years ago
6

By what percent must one increase the tension in a guitar string to change the speed of waves on the string from 328 m/s to 363

m/s?
Please show your work and the equations you used.
Physics
1 answer:
Elina [12.6K]4 years ago
4 0

Answer:

22.47 %

Explanation:

v 1 = 328 m/s, v 2 = 363 m/s

We know that the velocity of a wave in a stretch string is directly proportional to the square root of the tension in the string.

\frac{v1}{v2}=\sqrt{\frac{T1}{T2}}

\frac{T1}{T2}=\left ( \frac{328}{363} \right )^{2}

\frac{T1}{T2}=0.8165

Percentage increase in the tension

\frac{T2 - T1}{T1}\times 100 = \left (\frac{T2}{T1}-1  \right )\times 100

                                                  = \left ( \frac{1}{0.8165}-1 \right )\times 100

                                                 = 22.47 %

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3 years ago
our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure
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The coefficient of static friction between your partner and the floor is 0.55

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Mass m = 59 Kg

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From the formula of frictional force,

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
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To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).

By definition the efficiency of the heat engine is

\epsilon = 1- \frac{T_c}{T_h}

Where,

T_c = Temperature at the room

T_h  =Temperature of the soup

The work done is defined as,

dW = \epsilon(dQ_h)

Where Q_h represents the input heat and at the same time is defined as

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Where,

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The change at the work would be defined then as

dW = \epsilon(dQ_h)

dW = \epsilon c_v (dT_h)

dW = (1-\frac{T_c}{T_h})c_v (dT_h)

W = \int dW = \int (1-\frac{T_c}{T_h})c_v (dT_h)

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On the other hand we have that the heat lost by the soup is equal to

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The ratio between both would be,

\frac{W}{Q_h} = \frac{c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})}{c_v (T_h-T_c)}

\frac{W}{Q_h} = \frac{1+ln(\frac{T_h}{T_c})}{1-\frac{T_h}{T_c}}

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