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a_sh-v [17]
2 years ago
13

You accelerate a 0.43kg football 5m/s2. Calculate the force you applied to the football.

Physics
1 answer:
Elza [17]2 years ago
7 0

The force applied on the football is 2.15 Newton.

Given the data in the question;

  • Mass of football; m = 0.43kg
  • Acceleration; a = 5m/s^2

Force applied; F = \ ?

To determine the force applied on the football, we use Newton's laws of motion:

F = m * a

Where m is the mass of the object and a is the acceleration.

We substitute our given values into the equation

F = 0.43kg\ *\ 5m/s^2\\\\F = 2.15kg.m/s^2\\\\F = 2.15N

Therefore, the force applied on the football is 2.15 Newton.

Learn more: brainly.com/question/2388393

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Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

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2 years ago
A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),
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Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

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Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

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<u>FOR FORCE (P)</u>:

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<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

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<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

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