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ioda
3 years ago
7

A hammer taps on the end of a 4.00-m-long metal bar at room temperature. a microphone at the other end of the bar picks up two p

ulses of sound, one that travels through the metal and one that travels through the air. the pulses are separated in time by 1 1.0 ms. what is the speed of sound in this metal?
Physics
1 answer:
Nezavi [6.7K]3 years ago
4 0
<span>Length that is distance s = 4.00m
 Speed of sound in air V = 343 m/s
 Time for metal = 11.0ms = 0.011
 Calculating the time t = s / V = 4 / 343 = 0.01166s
 Time for the sound in the metal = 0.01166 - 0.011 = 0.00066 = 6.6 x 10^-4s Speed of sound in metal Vm = 4 / 6.6 x 10^-4 = 6.060 x 10^3 m/s</span>
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kodGreya [7K]

Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

E_i = E_f

so:

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where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2

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7 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

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Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

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3 years ago
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Answer:

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Einstein's laws of special relativity are only true in frames that move with contant speed to one another

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Answer:

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