<span>through friction between her feet and the carpet </span>
Force of earth gravity on 1 kg mass is 9.8 N.
Since the mass is at earth surface, therefore its distance from the centre of earth is equal to the radius of the earth i.e. 6400 km.
The force of earth gravity is calculated as
F=GMm/r^2 =(6.67*〖10〗^(-11)*6*〖10〗^24*1)/〖(64*〖10〗^6)〗^2 =9.8 N
Take 771.9 Kg M/s because that's how fast she is moving and divide it by 93 kg because that is her force pushing down
771.9/93= 8.3
Answer: A. 8.3 M/s
P.s I changed my answer on the last question you asked so change that.
Answer:
The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.
Explanation:
Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively
and the mass of big rock be 'M'
Initial momentum of the system equals

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'
Thus the final momentum of the system is

Equating initial and the final momenta we get

Now since the surface is frictionless thus the energy is also conserved thus

Similarly the final energy becomes
\
Equating initial and final energies we get

Solving i and ii we get

Using this in equation i we get
Thus putting v = -v' in equation i we get V' = 0
This implies Smaller stone rebounds while as larger stone remains stationary.
The net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.
To find the answer, we have to know more about the basic forces acting on a body.
<h3>How to find the net force on the box?</h3>
- Let us draw the free body diagram of the given box with the data's given in the question.
- From the diagram, we get,

where, N is the normal reaction, mg is the weight of the box,
is the net force, f is the kinetic friction.
- We have the expression for kinetic friction as,

- Thus, the net force will be,

Thus, we can conclude that, the net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.
Learn more about the basic forces on a body here:
brainly.com/question/28061293
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