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Snowcat [4.5K]
3 years ago
13

A woman falls to the ground while wearing a parachute. The air resistance on the parachute of the parachute is 500N. If the woma

n falls at a constant rate of 5m/s, then the gravitational force on her is
A) 500N
B) less than 500N
C) more than 500N
D) There is not enough information to draw a conclusion about the gravitation forces.
Physics
2 answers:
scoundrel [369]3 years ago
5 0

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, F_G, acting downward
  • The air resistance, F_D, 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:

\sum F=ma

where m is the mass of the woman and a her acceleration.

The net force can be written as

\sum F = F_G - F_D

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

a=0

Combining the equations together, we get:

F_G-F_D = 0

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

F_G=F_D=500 N

Learn more about forces and Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

inessss [21]3 years ago
5 0

Answer: A) 500N

Explanation: To be in equilibrium the weight must be equal to 500N to balance out air resistance.

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Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b
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Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

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After coming to a  flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

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U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

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After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

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