(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%.
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Answer:
66.24 Volts
Explanation:
W = Amount of work done in moving the charge from negative to positive terminal = 2.186 J
Q = Amount of charge being moved from negative to positive charge = 0.033 C
ΔV = EMF of the battery
Amount of work done in moving the charge from negative to positive terminal is given as
W = Q ΔV
2.186 = (0.033) ΔV
ΔV = 66.24 Volts
The answer is cooler. Hope this helps.
2.37eV stopping potential would be required to arrest the current of photoelectrons.
<h3 /><h3>What is stopping potential ?</h3>
The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximal kinetic energy of the electrons is equal to the stopping voltage.
Kmax = eV₀
2.37eV = eV₀
V₀ = 2.37eV
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