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castortr0y [4]
3 years ago
15

a horse moves a sleigh 1.00 kilometer by applying a horizontal 2,000 newton on its harness for 45 minutes.

Physics
1 answer:
sineoko [7]3 years ago
5 0
Work:

1 kilometer = 1000 meters
45 × 60 =  2700

W = F × D
W = 2,000 N × 1,000 m
W = 2,000,000 J

P = W ÷ t
P = 2,000,000 J ÷ 2,700 s
P = 741 watts 


Answer:

741 watts of horse power.
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The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string
Katyanochek1 [597]

Answer:

\dfrac{\mu_A}{\mu_G}=0.197

Explanation:

given,

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Speed on the G string = 128 m/s

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Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

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 v_A = \sqrt{\dfrac{F_A}{\mu_A}}.......(1)

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Assuming force is same in both the string

now,

dividing equation (2)/(1)

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

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• The net force in the parallel direction is

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• The net force in the perpendicular direction is

∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0

Solving the second equation for <em>n</em> gives

<em>n</em> = <em>mg</em> cos(21°)

<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)

<em>n</em> ≈ 1.83 N

Then the magnitude of friction is

<em>f</em> = <em>µn</em>

<em>f</em> = 0.25 (1.83 N)

<em>f</em> ≈ 0.457 N

Solve for the acceleration <em>a</em> :

-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

<em>a</em> ≈ -5.80 m/s²

so the block is decelerating with magnitude

<em>a</em> = 5.80 m/s²

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