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Tamiku [17]
3 years ago
7

How far does an object move if a 20 N force is applied to it doing 40 J of work?

Physics
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer: The object moved 2 metres away

Explanation:

Given that the

Force applied F = 20 N

Work done W = 40 J

From the definition of workdone which state that: product of force and distance

W.D = F × S

Where S = distance

S = W.D/F

S = 40/20 = 2 metres

The object moved 2 metres away.

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Two balls of equal size are dropped from the same height from the roof of a building. One ball has twice the mass of the other.
frutty [35]

Answer:

The kinetic energy of the more massive ball is greater by a factor of 2.

Explanation:

By conservation of energy, we know that the initial energy = final energy. At first, the balls are dropped from a height with no initial velocity so their initial energy is all potential energy. When they reach the bottom, all their energy is kinetic energy. So all of their energy is changed from potential to kinetic energy. This means that the ball with greater potential energy will have a greater kinetic energy.

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Additionally, let B1 = lighter ball with mass m and let B2 = heavier ball with mass m2. Since we know that intial potential energy = final kinetic energy. We can rewrite it as potential energy = kinetic energy = mass * gravity constant * height. For B1, it is mgh and for B2 it is 2mgh, so B2's kinetic energy is twice that of B1.

3 0
3 years ago
A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br
serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

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So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

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