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3 years ago

The amount of WORK done is determined by 2 factors.

Physics

1 answer:

RUDIKE [14]3 years ago

6 0

Explanation :

Work is done when a force is applied to create a displacement on an object.

Thus, the work done depends on the two factors i.e.

(1) Applied force (F)

(2) Distance or displacement (d)

Mathematically, work done is $W= F.s$

It also depends on the angle between the force and the displacement.

$W=Fs\ cos\ \theta$

For example,

A person carries a weight of 20 kg and lifts it on his head 1.5 m above the surface. So, the work done by him on the luggage will be:

$W= F\times s$

$W=m\times g \times s$

$W=20\ kg\times 9.8\ m/s^2\times 1.5\ m$

So, $W=294\ Joules$

Hence, the work done by him on the luggage is 294 Joules.

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Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 18 cm that is placed in a spatially uniform m

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Answer:

ε = 6.617 V

Explanation:

We are given;

Number of turns; N = 40 turns

Diameter;D = 18cm = 0.18m

magnetic field; B = 0.65 T

Time;t = 0.1 s

The formula for the induced electric field(E.M.F) is given by;

ε = |-NAB/t|

A is area

ε is induced electric field

While N,B and t remain as earlier described.

Area = π(d²/4) = π(0.18²/4) = 0.02545

Thus;

ε = |-40 × 0.02545 × 0.65/0.1|

ε = 6.617 V

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2 years ago

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Answer:

30 degrees

Explanation:

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See atached diagram

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Answer:

$331665750000\ m/s^2$

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Explanation:

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r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2$

The gravitational acceleration at the surface of such a star is $331665750000\ m/s^2$

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3 years ago