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Rudiy27
3 years ago
10

A copper rod at 25°C is 2.5 m long. How long would it take a sound to move through the rod from one end to another? how would I

do this?
Physics
2 answers:
andrew-mc [135]3 years ago
5 0
Length of the copper rod = 2.5 meters
Speed at which sound travels through copper = 3560 meter per second
Let us assume the time taken
by sound to cover the given distance = x seconds
We already know that 
Speed = Distance/ Time
Then
Time = Distance/ Speed
x = 2.5/3560 seconds
   = 0.0007 seconds.
This can ve done by hitting one end of the rod and then receiving the sound at the other end and using the stop clock to measure the time taken.
vladimir2022 [97]3 years ago
5 0

Answer;The time taken by they sound to move from one end of the rod to another end is 0.00054 seconds.

Explanation;

Speed of sound in the copper = 4600 m/s

Length of the copper  rod = 2.5 m

Speed=\frac{Distance}{\text{Time taken}}

Distance covered by the sound = length of the copper

4600 m/s=\frac{2.5 m}{T}

T=\frac{2.5 m}{4600 m/s}=0.00054 s

The time taken by they sound to move from one end of the rod to another end is 0.00054 seconds.

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Answer:

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For an object to move with constant velocity, the acceleration of the object must be zero:

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\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}.

So, the sum of the three forces must be zero:

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this implies:

\vec{F}_3  = - \vec{F}_1 - \vec{F}_2.

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector \hat{i} pointing east,  with the y axis unit vector \hat{j} pointing south, so the positive angle is south of east. For this, we got for the first force:

\vec{F}_1 = 83.7 \ N \ (-\hat{j}),

as is pointing north, and for the second force:

\vec{F}_2 = 59.9 \ N \ (-\hat{i}),

as is pointing west.

Now, our third force will be:

\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})

\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}

\vec{F}_3  =  (59.9 \ N , 83.7 \ N )

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To find the magnitude, we can use the Pythagorean theorem.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

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|\vec{F}_3| = 102.92 \ N

this is the magnitude.

To find the direction, we can use:

\theta = arctan(\frac{F_{3_y}}{F_{3_x}})

\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })

\theta = 57 \° 24 ' 48''

and this is the angle south of east.

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