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3 years ago

Prentice Hall. All

Physics

1 answer:

iVinArrow [24]3 years ago

7 0

Answer:

Decreases/Reduces

Explanation:

Fill in the blank:

Consider the equation Work = Force X Distance.

<em>If a machine increases the distance over which a force is exerted, the force </em>

<em>required to do a given amount of work</em> .........

If the work is a constant value, then by isolating force from the equation, we get:

Force = Work / Distance

By increasing the value of the Distance, then the quotient Work. Distance diminishes, and therefore the required force decreases (diminishes, reduces)

Answer: Decreases/Reduces

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Which of the following is not a subatomic particle?

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D. Nucleus because it is not a part of the group.

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A 24 kg crate is moving at a constant speed because it is being pushed with a force of 53n. What would be the coefficient of kin

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Answer:

The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.

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9 months ago

URGENTTT PLEASE HELPPPP. You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally i

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Answer:

Explanation:

heat lost by water will be used to increase the temperature of ice

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1 x 2090 x t

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Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

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a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

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$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

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where L is the length of the pencil.

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$\tau = I \alpha$ (1)

where

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Substituting into (1) and solving for $\alpha$ , we find:

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And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

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There are only two forces acting on the pencil here:

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- The normal reaction of the hand on the pencil, R

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$\tau = Fd$

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Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

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Answer:

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