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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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The answer would be C., Organic Molecule.
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The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.
oksano4ka [1.4K]

Answer:

i = 2.84 \times 10^{-3} A

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

J = \frac{di}{dA}

so the current through the wire is given as

i = \int J dA

now we have

i = \int_{0.921R}^R J dA

here we have

J = (3.25 \times 10^8)r^2

now plug in the values in above equation

i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr

now we have

i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr

i = (2.04 \times 10^9) \frac{r^4}{4}

now plug in both limits as mentioned

i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})

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i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)

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8 0
3 years ago
An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally
amid [387]

The tension in the supporting cable when the cab originally moves downward is 18422.4 N

What is tension?

Tension is described as the pulling force by the means of a three-dimensional object.

Tension might also be described as the action-reaction pair of forces acting at each end of said elements.

Here,

m =combined mass = 1600 kg

s = Displacement of the elevator = 42 m

g = Acceleration due to gravity = 9.81 m/s²

u = Initial velocity = 12 m/s

v = Final velocity = 0

According to the equation of motion:

v^{2} - u^{2} = 2as

0 - 12^2 = 2*a*42

a = - 144 / 84

a = - 1.714 m/s^2

Now let's write the equation of the forces acting on the elevator. Taking upward as positive direction:

T-mg = ma

T = m(g-a)

T = 1600 ( 9.8-(-1.74))

T=18422.4 N

Hence,

The tension in the supporting cable when the cab, originally moving downward is 18422.4 N

Learn more about tension here:

<u>brainly.com/question/13772148</u>

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The interpretation of the subject in question is characterized in the discussion section below.

Explanation:

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