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galben [10]
3 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]3 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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Answer:

m = 1.99 kg = 2 kg

Explanation:

The moment of inertia of a bicycle rim about it's center is given by the following formula:

I = mr^{2}\\

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r = 0.65 m/2 = 0.325 m

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Therefore,

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3 years ago
The magnitude of the Normal Force on a
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Answer:

5. Is greater than mg, always

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If the cone has an inclination of angle β, the sum of forces will be:

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y-axis:

N*cos β - m*g = m*ay   where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.

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N = \frac{m*g + m*ay}{cos \beta }

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7 0
3 years ago
A satellite is in a circular orbit 8200 km above the Earth’s surface; i.e., it moves on a circular path under the influence of n
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For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.

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Make V the subject of formula

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V² = AR

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V = √(1.87965 * 14.57x10^6)

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3 years ago
How do you find the normal force here? I forgot
kakasveta [241]
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3 years ago
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Answer:

a) 3673469.39 seconds

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Equation of motion

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