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ExtremeBDS [4]
3 years ago
6

The engine that moves the cables for the San Francisco cable cars delivers 390.3 kW of power for each line. How long does it tak

e for 6.5 ×10^6 J of work (about the amount of work needed to raise a partially loaded cable car up Nob Hill) to be done by this engine?
Physics
1 answer:
Musya8 [376]3 years ago
5 0

Answer:

It would take 16.7 s for the work to be done by the engine.

Explanation:

From the question, given: Power = 390.3 kW

                                           Work to be done = 6.5 x 10^{6} J

But, power and work done with respect to time, has a relationship of:

Power = \frac{work done}{time}

So that,

time = \frac{work done}{Power}

Thus,

time  = \frac{6.5*10^{6} }{390.3*10^{3} }

       = 16.6539

time = 16.7 s

Time required is 16.7 seconds.

Thus, it would take 16.7 s for the work to be done by the engine.

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An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q
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Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency f  = ( \dfrac{1}{2 \pi}) \omega

where;

\omega = \sqrt{\dfrac{k}{m} }

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{k}{m} }  \Bigg )

However;

k  = \dfrac{F}{x} and;

mass m = m_{car } + m_{person}

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} }  \Bigg )

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} }  \Bigg )

where;

F = m_{person}g

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} }  \Bigg )

replacing the values;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} }  \Bigg )

\mathbf{f = 0.442 \ Hz}

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3 years ago
This is for physical science. Can someone help me?
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A. 320 g
B. 160 g
C. 80 g
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6 0
3 years ago
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All moving objects don’t have momentum<br> A. True<br> B. False
mestny [16]

Answer:

No

Explanation:

Some objects gain momentum.

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A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

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Explanation:

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