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2 years ago

A book weighing 2.0 Newtons is lifted 3.0 meters in 4.0 seconds. How much work was done? SHOW WORK

Physics

1 answer:

qaws [65]2 years ago

7 0

6 Ns is the work done on the book to lift it by 3 m.

Answer:

Explanation:

Work done is the measure of amount of force required to move an object from its initial position to any desired position. So mathematically it can be calculated as the product of force acting on that object to the displacement of that object. Thus, work done on any object or by any object is directly proportional to the force acting on that object and the displacement of that object.

Work done = Force × displacement

As here the force is said to be 2 N and the displacement of the book is said to be 3 m, the work done on the book is

Work done = 2×3 = 6 N.s.

The time has no role to do on the work done calculation to be performed on the object.

Thus, 6 Ns is the work done on the book to lift it by 3 m.

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During a compression at a constant pressure of 290 Pa, the volume of an ideal gas decreases from 0.62 m3 to 0.21 m3. The initial

Aloiza [94]

Answer:

a) -41.1 Joule

b) 108.38 Kelvin

Explanation:

Pressure = P = 290 Pa

Initial volume of gas = V₁ = 0.62 m³

Final volume of gas = V₂ = 0.21 m³

Initial temperature of gas = T₁ = 320 K

Heat loss = Q = -160 J

Work done = PΔV

⇒Work done = 290×(0.21-0.62)

⇒Work done = -118.9 J

a) Change in internal energy = Heat - Work

ΔU = -160 -(-118.9)

⇒ΔU = -41.1 J

∴ Change in internal energy is -41.1 J

b) V₁/V₂ = T₁/T₂

⇒T₂ = T₁V₂/V₁

⇒T₂ = 320×0.21/0.62

⇒T₂ = 108.38 K

∴ Final temperature of the gas is 108.38 Kelvin

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3 years ago

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

$-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}$

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

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