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3 years ago

A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight

Physics

1 answer:

tankabanditka [31]3 years ago

3 0

Answer:

There is no work done.

Explanation:

Given the following data;

Mass = 10 kg

To find the work done?

In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.

Mathematically, work done is given by the formula;

Work done = force * distance

$W = F * d$

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.

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olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

No. of electrons in the given amount of charge:

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

Now the no. of moles in this many molecules:

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where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

So, the mass of water in the obtained moles:

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

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Read 2 more answers

A sledge loaded with bricks has a total mass of 18.3 kg and is pulled at constant speed by a rope inclined at 19.8° above the ho

vladimir1956 [14]

Answer

given,

total mass = 18.3 Kg

inclined at = 19.8°

distance = 19.7 m

coefficient of friction = 0.5

a) Horizontal component of force forward,

$F_x = F cos \theta$

Horizontal force backward = force of friction, f =μN.

Normal force, N = mg - F sinθ

Since the velocity is uniform, acceleration, a = 0

Net force, ΣF = 0

F cos θ+ -μ(mg -F sinθ) = 0

F (cosθ + μsin θ) = μmg

$F = \dfrac{\mu m g}{cos \theta + \mu sin \theta}$

$F = \dfrac{0.5\times 18.3\times 9.8}{cos 19.8^0 +0.5\times sin 19.8^0}$

=80.76 N

(b) Work done by the rope on the sledge,

W = F.S cosθ = 80.76 x 19.7 x 0.9360

= 1497 J

E = W done against friction,

W' = f x S =F x S = 80.76 x 19.7 x0.9360

= 1497 J

we know velocity is uniform,.

W done by the applied force = W done against friction.

= 1440 J

3 0

3 years ago