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3 years ago

A 0.72-m long string has a mass of 4.2 g. The string is under a tension of 84.1 N. What is the speed of a wave on this string?

Physics

1 answer:

frosja888 [35]3 years ago

5 0

The speed of the wave in the string is 83.4 m/s

Explanation:

For a standing wave in a string, the speed of the wave is given by the equation:

$v=\frac{1}{2L}\sqrt{\frac{T}{m/L}}$

where

L is the length of the string

T is the tension in the string

m is the mass of the string

In this problem, we have:

L = 0.72 m

m = 4.2 g = 0.0042 kg

T = 84.1 N

Solving the equation, we find the speed of the wave:

$v=\frac{1}{2(0.72)}\sqrt{\frac{84.1}{0.0042/0.72}}=83.4 m/s$

Learn more about waves:

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Answer:

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a) Nt=25stitches per sec

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<h3>Number of stitches per sec and he mass of oscillation motion</h3>

Question Parameters:

This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.

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Generally the equation for the Number of stitches per sec is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

Generally the equation for the Time t is mathematically given as

$T=2\pi\sqrt{\frac{m}{k}}$

Therefore

$0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}$

m=2.033e-5kg

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2 years ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

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(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

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And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

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In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

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3 years ago

How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams

Harrizon [31]

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

H = mL

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Insert the parameters and solve;

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H = 45200J

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2 years ago