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3 years ago

14

A 60 kg skier starts from rest at the top of a frictionless slope of height of 35 meters. The velocity at the bottom is v. If a

helium balloon is tied to the skier and pulls directly up (vertical) with a force of 200 N, what is the skier's speed at the bottom of the hill compared to the speed without the parachute (v)

2 answers:

Solnce55 [7]3 years ago

8 0

Answer:

Explanation:

Speed of skier without parachute

= √ 2gh

= √ 2 x 9.8 x 35

= 26.2 m / s

Speed of skier with parachute

net force downwards

mg - 200

= 60 x 9.8 -200

= 388 N

acceleration = 388 / 60

a = 6.47 m / s

v = √ 2ah

= √ 2 x 6.47 x 35

= 21.28 m / s

Ksju [112]3 years ago

6 0

Answer:

Explanation:

mass, m = 60 kg

distance, h = 35 m

Force, f = 200 N

When the parachute is not used

let the velocity is v.

Use third equation of motion

v² = u² + 2gh

v² = 2 x 9.8 x 35

v = 26.2 m/s

When the parachute is used:

Let the velocity is v'

Use work energy theorem

mgh - F x h = 0.5 mv'²

60 x 9.8 x 35 - 200 x 35 = 0.5 x 60 x v'²

20580 - 7000 = 30 v'²

v' = 21.3 m/s

So, the ratio of v' : v = 21.3 : 26.2 = 0.8

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$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

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$\theta = 7.46\,rev$

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