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Ulleksa [173]
1 year ago
15

Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating porti

on of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?
Physics
1 answer:
mylen [45]1 year ago
6 0

(a) The tension the musician must stretch it is 147.82 N.

(b) The percent increase in tension is needed to increase the frequency is 26%.

<h3>Tension in the string</h3>

v = √T/μ

where;

  • v is speed of the wave
  • T is tension
  • μ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/m

v = Fλ

in fundamental mode, v = F(2L)

v = 2FL

v = 2 x 65.4 x 0.6 = 78.48 m/s

v = √T/μ

v² = T/μ

T = μv²

T = 0.024 x (78.48)²

T = 147.82 N

<h3>When the frequency is 73.4 Hz;</h3>

v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s

T = μv²

T = (0.02)(88.08)²

T = 186.19 N

<h3>Increase in the tension</h3>

= (186.19 - 147.82)/(147.82)

= 0.26

= 0.26 x 100%

= 26 %

Thus, the tension the musician must stretch it is 147.82 N.

The percent increase in tension is needed to increase the frequency is 26%.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

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#82

here we know that

acceleration = 2 m/s/s

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#83

here we know that

acceleration = 3 m/s/s

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now we can use kinematics to find the final speed

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#84

here we know that

acceleration = 7 m/s/s

time = 3 s

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now we can use kinematics to find the final speed

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6 0
3 years ago
A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is
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Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

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mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

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final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m

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Olegator [25]

Answer:

1) Addition of a catalyst

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