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antiseptic1488 [7]
3 years ago
10

When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

d terminal speed: a. his speed is equal to g. b. the force of air drag on him is equal to his weight. c. the force of air drag on him is equal to zero. d. his acceleration is equal to g. e. the force of air drag on him is equal to g.
Physics
1 answer:
Masteriza [31]3 years ago
3 0

Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

      (G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

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Answer:

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Explanation:

We are given;

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Refractive index of benzene;n = 1.5

Now, let's calculate the wavelength of the film;

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Thus; λ_film = 585 x 10^(-9)/1.5

λ_film = 39 x 10^(-8) m

Now, to find the thickness, we'll use the formula;

2t = ½m(λ_film)

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Thus;

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Divide both sides by 2 to give;

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3 years ago
A 3600 kg rocket traveling at 2900 m/s is moving freely through space on a journey to the moon. The ground controllers find that
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Answer:

m=417.24 kg

Explanation:  

Given Data

Initial mass of rocket  M = 3600 Kg

Initial velocity of rocket vi = 2900 m/s  

velocity of gas vg = 4300  m/s

Θ = 11° angle in degrees

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Solution

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Vi=563.7 m/s

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m = (M ×vi ×tanΘ)/( vg + vi tanΘ)

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m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)

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4 years ago
Suppose a moving car has 2000 J of kinetic energy. If the carʹs speed doubles, how much kinetic energy would it then have?
Bond [772]

Answer:

option A

Explanation:

given,

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we know,

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now, speed of the car is doubled

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from equation (1)

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KE_2 = 4\times 2000

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Answer:????

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