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e-lub [12.9K]
1 year ago
8

A 5.0-kg object is suspended by a string from the ceiling of an elevator that is accelerating downward at a rate of 2.6 m/s 2. W

hat is the tension in the string
Physics
1 answer:
Tcecarenko [31]1 year ago
8 0

The tension in the string is 36 N.

To find the answer, we need to know about pseudo acceleration.

<h3>What's pseudo acceleration?</h3>
  • When an object is placed on a vehicle, it experiences a pseudo acceleration in the opposite direction of the motion of the vehicle.
  • Magnitude of the pseudo acceleration is same as that of the acceleration of vehicle.
<h3>What's the pseudo acceleration of a mass suspended in a lift that's moving downward with an acceleration of 2.6 m/s²?</h3>
  • Here, the acceleration of the lift is 2.6m/s², so the mass suspended by a string inside it will feel a pseudo acceleration of 2.6m/s² upward.
  • So the effective acceleration of the mass = acceleration due to gravity - pseudo acceleration = 9.8 - 2.6 = 7.2 m/s²
<h3>What's the tension of the string suspending the mass?</h3>

Tension of the string= mass × effective acceleration

= 5 × 7.2 = 36 N

Thus, we can conclude that the tension in the string is 36 N.

Learn more about the pseudo acceleration here:

brainly.com/question/26644944

#SPJ4

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Exercises
Crank

\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}

\\ \rm\Rrightarrow u=\dfrac{190}{-14}

\\ \rm\Rrightarrow u=13.6cm

Real

5 0
2 years ago
A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.
Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

P = \frac{W}{t}

W= F*s

Therefore,

P=\frac{F*s}{t}

Substitute the given values of force , displacement and time

F = 124 N,s = 32.0 m,t = 8.0 s

P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W

Thus the Power can be rounded off to the nearest value of 500 W

3 0
3 years ago
Read 2 more answers
Someone please help with this question
Alex787 [66]
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5 0
3 years ago
If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
AleksAgata [21]

Answer:

a) f_1=5.587Hz

b) f_{n+1}-f_n=5.587Hz

Explanation:

The frequency of the n^{th} harmonic of a vibrating string of length <em>L, </em>linear density \mu under a tension <em>T</em> is given by the formula:

f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

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3 years ago
Rank the ten objects from loudest to softest.
frez [133]
Um what are the ten objects..?
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