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Natali [406]
3 years ago
13

Luz, who is skydiving, is traveling at terminal velocity with her body parallel to the ground. She then changes her body positio

n to feet first toward the ground. What happens to her motion? She will continue to fall at the same terminal velocity because gravity has not changed. She will slow down because the air resistance will increase and be greater than gravity. She will speed up because air resistance will decrease and be less than gravity. She will begin to fall in free fall because she will have no air resistance acting on her.
Physics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Answer:

Option C - she will speed up because air resistance has reduced and be less than gravity

Explanation:

We are told that Luz is skydiving with terminal velocity and her body parallel to the ground. Now, at this point she will be experiencing a gravitational force acting downwards, and also air resistance as a result of the drag force on her body

Since the downward gravitational force on Luz is constant, she will fall with a net force of;

F_net = F_g - F_d

where;

F_net is the net force on Luz acting downwards

F_g is the gravitational force on Luz

F_d is the drag force on Luz

The drag force on her body is proportional to the surface area of attack.

We are now told that Luz changes her body position to feet first toward the ground. This means that the surface area of attack is reduced because the feet will consume less space than the frontal part of her body. Thus, the drag force will be lesser then before she changed her body position due to reduced air resistance on her body.

Now, from earlier, we saw that;

F_net = F_g - F_d

So, the lesser F_d is, the higher F_net becomes.

Thus, she will speed up because air resistance has reduced and be less than gravity.

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During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
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Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

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3 years ago
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