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Anni [7]
4 years ago
15

While skiing in Jackson, Wyoming, your

Physics
1 answer:
grin007 [14]4 years ago
8 0
Hello!

Work is equal to negative change in PE and PE=mgh and Power is change in work/change in time

So we have 

56.1kg * 9.80m/s^2 * 16m = 8796.5 J of PE at top of run and -8796.5J of work done also

if it takes him 12s then we need to find the time gravity alone would bring him to the bottom. we get

16m=1/2at^2 and solving for t we get 
t=sqrt(32/9.80m/s^2)=1.81s

so the difference in time is 12-1.8=11.2s

now we can find avg power=work/time = 8796.5J/11.2s = 785W or 785J/s

Hope this helps! Any questions please ask! Thank you!
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a bathtub full of warm water

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A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure
Nesterboy [21]

Answer:

<h2>The work done is 0.882 Joules.</h2>

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7 0
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4 years ago
A 60 W light bulb has a resistance of 880 Ω
Andreas93 [3]

<h2>\bf{ \underline{Given:- }}</h2>

\sf{• \:  Power \:  (P) = 60 \:  w}

\sf{•  \: Resistance \:  (R) = 880  \: Ω}

\\

<h2>\bf{ \underline{To \:  Find :- }}</h2>

\sf• \:  The  \: Current \:  (I).

\\

\huge\bf{ \underline{ Solution :- }}

\sf We  \: know  \: that,

\bf \red {\bigstar{ \: P = I^{2} R}}

\rightarrow \sf 60 = I^{2}  \times 880

\rightarrow \sf  \frac{60}{880} = I^{2}

\rightarrow \sf  I^{2}   = 0.0681

\rightarrow \sf  I   =  \sqrt{0.0681}

\rightarrow \sf  I   =  0.261 \:  \: (approx.)

\\

\bf The \:   \: value \:  \:  of \:  \:  I  \:  \:   is \:  \:  0.261 \:  \: Amps. \:  \: (approx.)

7 0
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