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worty [1.4K]
3 years ago
15

5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be

changed to raise the wave speed to 180 m/s?
Physics
1 answer:
ANTONII [103]3 years ago
3 0

Answer:

The tension on string when the speed was raised is 134.53 N

Explanation:

Given;

Tension on the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The speed of the wave is given as;

v = \sqrt{\frac{T}{\mu} }

where;

μ is mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2}  = \frac{T_2}{v_2^2}

The final tension T₂ will be calculated as;

T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N

Therefore, the tension on string when the speed was raised is 134.53 N

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Voltage needed to raise current to 3.75a using 20,20,200 resistor set
Varvara68 [4.7K]

<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

We are given:

R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega

Using above equation, we get:

\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega

Calculating the voltage by using Ohm's law:

V=IR         .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = 9.52\Omega

Putting values in equation 1, we get:

V=3.75\times 9.52\\\\V=35.7V

Hence, the voltage needed is 35.7 V

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A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 670 cm2. It is filled with oil of density 510 k
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Explanation:

Given

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