Physics, 25.06.2019 14:50 aurorasweetpea610

A64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. she then moves in a circle of radius 0.890 m around the pole. (a) determine the magnitude of the force exerted by the horizontal rope on her arms. kn (b) compare this force with her weight. frope w =

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We have that for the Question it can be said that the magnitude of the force exerted by the horizontal rope on her arms and the ratio of the Force to the weight is

F=1150.561NF/W=1.8325

From the question we are told

A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. F-rope W =

Generally the equation for the force applied is mathematically given as

$F=\frac{( mv^2)}{R}\\\\Therefore\\\\F=\frac{( mv^2)}{R}\\\\F=\frac{( (64)(4.0)^2)}{0.890}\\\\$

F=1150.561N

Generally the equation for the Weight is mathematically given as

W=mg

Therefore

W=64*9.81

W=627.84N

Therefore

The Force to weight ratio is

F/W=1150.561N/627.84N

F/W=1.8325

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Answer from: sanchez7489

(a) 1173.7 N

The tension in the horizontal rope provides the centripetal force that keeps the ice skater in circular motion around the pole:

$T=m\frac{v^2}{r}$

where

m = 64.0 kg is the mass of the ice skater

v = 4.04 m/s is the speed of the ice skater

r = 0.890 m is the radius of the circular trajectory

T is the tension in the rope

Solving for T, we find

$T=(64.0 kg)\frac{(4.04 m/s)^2}{0.890 m}=1173.7 N$

(b) The tension in the rope is 1.87 times larger than the weight of the ice skater

The weight of a person is given by:

W = mg

where

m is the mass of the person

g = 9.8 m/s^2 is the acceleration due to gravity

For the ice skater in this problem,

m = 64.0 kg

so her weight is

W=(64.0 kg)(9.8 m/s^2)=627.2 N

So we see that the ratio between the tension in the rope and the weight of the person is

$\frac{T}{W}=\frac{1173.7 N}{627.2 N}=1.87$

Answer from: Quest

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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