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galben [10]
4 years ago
10

A sled is dragged along a horizontal path at a constant speed of 1.50 m/s by a rope that is inclined at an angle of 30.0° with r

espect to the horizontal. The total weight of the sled is 470 N. The tension in the rope is 230 N. How much work is done by the rope on the sled in a time interval of 10.0 s?
Physics
1 answer:
Allisa [31]4 years ago
4 0

Answer:2987.7 J

Explanation:

Given

sled weight is W=470 N

Tension in rope T=230 N

Time interval t=10 s

velocity of sled v=1.5 m/s

here speed is equal to velocity which is constant therefore acceleration is zero

thus F_{net}=0

Distance traveled in t=10 s

x=v\times t=1.5\times 10=15 m

considering surface to be friction less

thus T\cos 30 will do work on sled

W=F\cdot x

W=T\cos 30\cdot x

W=230\cos 30\times 15

W=2987.78 J\approx 2.987 kJ

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<span>a) 1960 m b) 960 m Assumptions. 1. Ignore air resistance. 2. Gravity is 9.80 m/s^2 For the situation where the balloon was stationary, the equation for the distance the bottle fell is d = 1/2 AT^2 d = 1/2 9.80 m/s^2 (20s)^2 d = 4.9 m/s^2 * 400 s^2 d = 4.9 * 400 m d = 1960 m For situation b, the equation is quite similar except we need to account for the initial velocity of the bottle. We can either assume that the acceleration for gravity is negative, or that the initial velocity is negative. We just need to make certain that the two effects (falling due to acceleration from gravity) and (climbing due to initial acceleration) counteract each other. So the formula becomes d = 1/2 9.80 m/s^2 (20s)^2 - 50 m/s * T d = 1/2 9.80 m/s^2 (20s)^2 - 50m/s *20s d = 4.9 m/s^2 * 400 s^2 - 1000 m d = 4.9 * 400 m - 1000 m d = 1960 m - 1000 m d = 960 m</span>
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Comic strip hero superman meets an asteroid in outer space and hurls it at 100m/s. he has a mass of "m". The astroid has a mass
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