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torisob [31]
3 years ago
14

A light woman and a heavy man jump from an airplane at the same time and open their same-size parachutes at the same time. which

person will get to a state of zero acceleration first?
Physics
2 answers:
Hitman42 [59]3 years ago
7 0

Answer:

two people if they have a similar area should reach zero acceleration at the same time.

Explanation:

Let's analyze what is the acceleration of each person using Newton's second law, before opening the parachute

    mg = ma   ⇒   a = g

Therefore, the two people have the same acceleration, we assume that the size (area) of the two is not very different, so the speed of the two people should be the same at all times.

Now let's analyze what happens when opening the parachute, this creates a force of resistance to the fall, which is proportional to the area of ​​the parachute , that is always much larger than the area of ​​people and the speed of the bodies,

     fr -mg = ma

As we can see the difference in acceleration in each person should be in the resistance force of the parachute, since from the previous part we saw that the speeds of the two people are equal and all the parachutes have the same area, the force of friction should be the same for both people, this force is increasing until the acceleration is zero and from there on we have the terminal speed.

In conclusion, from the above the two people if they have a similar area should reach zero acceleration at the same time.

Airida [17]3 years ago
6 0
The heavy man will fall faster, because he has more mass and heavy things fall faster than light things
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The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o
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Answer:

time taken is equal to 14,156 years

Explanation:

we know,

Y=Ae^{-kt}

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\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}

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5 0
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When you blink your eye, the upper lid goes from rest with your eye open to completely covering your eye in a time of 0.024 s.
dusya [7]

a) Distance moved by the top lid during a blink: 1 cm (estimate)

b) The acceleration is 34.7 m/s^2

c) The final speed is 0.83 m/s

Explanation:

a)

For the purpose of this problem, we can estimate the size of the eye from the top lid to the bottom lid to be 1 cm, so this is the distance moved by the top lid during a blink.

b)

Assuming the motion of the eyelid to be at constant acceleration, we can find the acceleration by using the following suvat equation:

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For the eyelid, we have:

u = 0 (it starts from rest)

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t = 0.024 s is the time taken

Solving for a, we find the acceleration:

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v is the final speed

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u = 0

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t = 0.024 s

Substituting, we find the final speed:

v=0+(34.7)(0.024)=0.83 m/s

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