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3 years ago

How much work do you do on a 15 N book in lifting it straight up for a distance of 0.40 meters?

Physics

1 answer:

astra-53 [7]3 years ago

7 0

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of 0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

$W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J$

So, the magnitude of work done in lifting the book is 6 joules.

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Can a light bulb work without energy transfer.

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Answer:Energy cannot be created or destroyed. It can be stored, or it can be transferred i.e. from a hot object to a cool object by conduction, convection or radiation.

Explanation:

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3 years ago

If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p

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The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

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By Newton's second law, the magnitude of the force F is proportional to the acceleration a according to

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3 years ago

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I have 2 bulbs one coverts 60% of energy to light 45% which is the most efficient and what happens to the rest of the energy

Leya [2.2K]

Answer:

The first one (60%)

Explanation:

The first one converts 15% more energy than the other one. Therefore, it is more efficient.

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3 years ago

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A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

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3 years ago