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3 years ago

Please Help!.....................

Physics

1 answer:

Setler79 [48]3 years ago

4 0

The awnser is tranverse wave
hope that helps

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A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma

goldfiish [28.3K]

(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

$F=ma$

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2$

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

$F=ma$

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

$a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2$

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

$x(t)=\frac{1}{2}a_g t^2$

where $a_g = 0.098 m/s^2$ is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

$x'(t)=15-\frac{1}{2}a_s t^2$

where $a_s=0.8 m/s^2$ is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.