The gravitational force on the woman is A) 500 N
Explanation:
There are two forces acting on the woman during her fall:
- The force of gravity,
, acting downward - The air resistance,
, acting upward
According to Newton's second law, the net force acting on the woman is equal to the product between the woman's mass and her acceleration:

where m is the mass of the woman and a her acceleration.
The net force can be written as

Also, we know that the woman falls at a constant velocity (5 m/s), this means that her acceleration is zero:

Combining the equations together, we get:

which means that the magnitude of the gravitational force is equal to the magnitude of the air resistance:

Learn more about forces and Newton's second law:
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Answer:
he fall movement we see that both the force is different from zero, and the torque is different from zero.
When analyzing the statements the d is true
Explanation:
Let's pose the solution of this problem, to be able to analyze the firm affirmations.
When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.
The equation for this part is
Fe - W = m a
k x - mg = m a
As the axis of rotation is located at the top where they jump, there is a torque.
What is it
Fe y - W y = I α
angular and linear acceleration are related
a = α r
Fe y - W y = I a / r
In the fall movement we see that both the force is different from zero, and the torque is different from zero.
When analyzing the statements the d is true
No cluuuueee :/ sowwwwyyy but good luck
E=energy=5.09x10^5J = 509KJ
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
<span>two filters with vertical polarization</span>