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masya89 [10]
3 years ago
15

What horizontally-applied force will accelerate a crate of mass 400 kg at 1 meter per second per second across a factory floor a

gainst a friction force of half its weight?
Physics
1 answer:
son4ous [18]3 years ago
6 0

Answer:

The horizontally applied force = 2360 N

Explanation:

<em>Force:</em> Force can be defined as the product of mass and acceleration. the S.I unit of force is Newton (N)

Fh = Fr + ma......... Equation 1

Where Fh = horizontally applied force, Fr = friction force, m = mass of the crate, a = acceleration of the crate.

<em>Given: m = 400 kg, a = 1 m/s²</em>

Fr = 1/2 W, W = mg  ⇒W = 400×9.8 = 3920 N

∴Fr = 1/2(3920), Fr = 1960 N

Substituting these values into equation 1

Fh = 1960 + 400×1

Fh = 1960 + 400

Fh = 2360 N

Therefore the horizontally applied force = 2360 N

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assuming that there is an equal amount of each base, you can make 65,536 bases

Explanation:

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Two marbles, one twice as heavy as the other, are dropped to the ground from the roof of a building. Just before hitting the gro
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Answer:

 B. twice as much kinetic energy

Explanation:

Lets take the mass of the first marble =2 m

the mass of the second marble = m

We know that velocity of particle does not depends on their mass that is the velocity of both mass will be same after dropping from the roof.

We know that kinetic energy of a mass is given as

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Kinetic energy for light mass

KE'=\dfrac{1}{2}\times m\times V^2

KE=2 KE '

Form above two equation we can say that ,the kinetic energy for the heavier mass is twice the lighter mass.

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What contributions did J.J. Thomson make to atomic history?
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If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?
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Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

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