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Ray Of Light [21]
2 years ago
9

A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak

e a pulse to travel from one support to the other?
Physics
1 answer:
MaRussiya [10]2 years ago
5 0
Ans: Time <span>taken by a pulse to travel from one support to the other = 0.348s
</span>
Explanation:
First you need to find out the speed of the wave.

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

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = \sqrt{ \frac{150}{0.0232} } = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s
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Zoroastrianism affected the way the Persians governed their subjects by allowing the territories under their rule to worship their own religion.
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If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di
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Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

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3 years ago
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did
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Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

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It divides:

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How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

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3 years ago
9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.
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Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

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where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

W = 50*9.8*13 = 6370 J

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