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11111nata11111 [884]
1 year ago
5

Q10. In an ideal situation without any air resistance, any free-falling body will fall at the same rate of acceleration, regardl

ess of its mass, and thus would simultaneously hit the
ground if released from the same height. However, in real-life scenarios, air resistance plays a vital role in determining the fall rate of any object. Which parameters would affect
the fall if air drag is also considered? Choose all possible answers.
Weight of the object.
>>
Shape of the object.
Density of the air.
Velocity of the object.
Submit
Physics
1 answer:
rosijanka [135]1 year ago
7 0

Velocity of the object parameters would affect the fall if air drag is also considered.

<h3>What is velocity in physics with example?</h3>

Velocity can be defined as the rate at which that moves in a specific direction. as the velocity of a car driving north on a road or the pace at which a rocket takes off. Because the velocity vector is scalar, its integrand magnitude will always equal the motion's speed.

<h3>Can velocity be negative?</h3>

A moving object has a negative velocity when it is moving in the wrong direction. If an object is slowing down, its generate and evaluate is pointing away from the direction in which it is moving (in this case, a positive acceleration).

<h3>What is the principle of velocity?</h3>

An object must move at a constant speed and direction in order to have a constant velocity. The object can only travel in a straight line if the direction is constant. So, motion in a single direction at a constant speed is defined as having a constant velocity.

To know more about Velocity visit:

brainly.com/question/28738284

#SPJ13

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<u>Explanation:</u>

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4 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
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Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
4 years ago
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How can you verify the archimedes principle?​
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Call the applied force 'A'. (Clever ?)

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The net force on the art is (A-15) forward.

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A - 15 = 38.115

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