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3 years ago

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A ski gondola is connected to the top of a hill by a steel cable of length 600 m and diameter 1.2 cm . As the gondola comes to t

he end of its run, it bumps into the terminal and sends a wave pulse along the cable. It is observed that it took 18 s for the pulse to return. Part A What is the speed of the pulse?

1 answer:

Gemiola [76]3 years ago

6 0

Answer:

v = 66.7 m/s

Explanation:

Given that,

The length of steel cable, L = 600 m

Diameter = 1.2 cm

It is observed that it took 18 s for the pulse to return.

The time taken to cover 600 m will be :

t = T/2

t = 9 s

Let v be the of the pulse. We know that,

$v=\dfrac{L}{t}\\\\v=\dfrac{600}{9}\\\\v=66.7\ m/s$

So, the speed of the pulse is equal to 66.7 m/s.

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$\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 1/4 \ mile = 402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}$

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2 years ago

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Goshia [24]

Answer:

The rate of heat removed from inside the refrigerator is 300 watts.

Explanation:

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$\dot Q_{L} = \dot Q_{H}-\dot W$ (1)

Where:

$\dot Q_{H}$ - Rate of heat released to the room, in watts.

$\dot W$ - Rate of electric energy needed by the refrigerator, in watts.

If we know that $\dot Q_{H} = 800\,W$ and $\dot W = 500\,W$ , then the rate of heat removed from inside the refrigerator is:

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The rate of heat removed from inside the refrigerator is 300 watts.

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