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Wittaler [7]
3 years ago
5

A 2.31-kg rope is stretched between supports that are 10.4 m apart, and has a tension in it of 49.2 N. If one end of the rope is

slightly tweaked, how long will it take for the resulting disturbance to reach the other end?
Physics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

The resulting disturbance takes approximately 0.70 seconds to reach the other end.

Explanation:

The velocity of a disturbance on a rope depends on the physical properties of the material and the tension exerted to it, the equation is:

v=\sqrt{\frac{T}{\mu}}  (1)

With v the velocity of the disturbance, T the tension of the rope and μ the linear density of the rope. The linear density of the rope is the relation between mass M and length L so we have:

\frac{M}{L} (2)

Using (2) on (1):

v=\sqrt{\frac{TL}{M}}=\sqrt{\frac{49.2*10.4}{2.31}}\simeq14.9\frac{m}{s} (3)

If we assume the velocity of the disturbance is constant, it will take for the resulting disturbance to reach the other ends a time t:

t=\frac{L}{v}=\frac{10.4}{14.9}\simeq0.70\,s (4)

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kramer

Answer:

60.2 J

Explanation:

Efficiency is the ratio of work out to work in.

e = Wout / Win

0.86 = Wout / 70 J

Wout = 60.2 J

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What difference will you get from the flying or air filled balloon and the hydrogen filled balloon.​
mash [69]

Answer:

the hydrogen filled balloon is denser than the air filled balloon

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What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300
Vinvika [58]
The efficiency of an ideal Carnot heat engine can be written as:
\eta = 1-  \frac{T_{cold}}{T_{hot}}
where
T_{cold} is the temperature of the cold region
T_{hot} is the temperature of the hot region

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4 0
3 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

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