All motion will eventually stop due to the "loss" of energy to__

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

or

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

or

$\eta=0.52$

or

Efficiency = 52%

8 0

4 years ago

How much work does it take to lift an object weighing 200N to a distance of 20 meters

Bogdan [553]

Answer:

4000 J

Explanation:

W = F = 2000N , s = 20 m

Workdone = F × s = 200 × 20 = 4000J

7 0

3 years ago

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A tall cylinder with a cross-sectional area 12.0 cm² is partially filled with mercury; the surface of the mercury is 5.00 cm abo

JulijaS [17]

Answer:

$V = 816 cm^3$

Explanation:

As we know that gauge pressure of the fluid at the bottom of the cylinder is given as

$P = \rho g h$

now we know that pressure at the bottom is double when water is poured on the mercury

So we have

$\rho_{hg} g h_1 = \rho_w g h_2$

so we will have

$13.6 \times 5 = 1 \times h$

so we have

$h = 68 cm$

now the volume of the water added to it is given as

$V = h A$

$V = (68 cm)(12 cm^2)$

$V = 816 cm^3$

6 0

4 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 55.0°-angle with the horizontal. (a

Nikitich [7]

Answer:

Check attachment for the free body diagram

Explanation:

Given that,

Ladder length L = 14m

Weight of ladder W= 490N

The weight will act at the midpoint

i.e at 14/2 = 7m, L1 = 7m

The ladder makes an angle of 55° with the horizontal. θ=55°

Weight of firefighter Wf =810N

The firefighter is at 3.9m from the horizontal ground, L2 =3.9

The wall exerts a force on the ladder, let It be Nw

The ground exerts a force on the ladder, let it be Ng

The let Ff be the frictional force that opposes motion.

a. We want to find the horizontal and vertical force the ground exerted on the ladder i.e Ng and Ff.

Using Newton second law

ΣFy= m•ay

ay=0, since the body is not accelerating

ΣFy = 0

Ng — Wf —W = 0

Ng = Wf + W

Ng = 490 + 810

Ng = 1300 N.

Also,

ΣFx= m•ax

ax=0, since the body is not accelerating

ΣFx = 0

Ff — Nw = 0

Ff = Nw

Now,

Let take moment about point A(ground), but note before we take moment, the forces must be perpendicular to the ladder.

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•3.9 + (490Cos55)•7 = (NwSin55)•14

1811.93 + 1967.37 = 11.47Nw

3779.3 = 11.47Nw

Then, Nw = 3779.3/11.47

Nw= 329.49N

Since Nw = Ff

Then Ff = 329.49N

So the required reaction exerted by the ground on the ladder are

Ng = 1300 N

Ff = 329.49 N

b. Now the firefighter is a distance of 9.4m from the horizontal and the ladder is about to slip

So we need to calculate the coefficient of static friction μs

Check attachment for new diagram,

So calculating for Nw again, since the firefighter have new position

Now the firefighter is at 9.4m from the ground. Therefore, L2 = 9.4m

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•9.4 + (490Cos55)•7 = (NwSin55)•14

4367.21 + 1967.37 = 11.47Nw

6334.58 = 11.47Nw

Then, Nw = 6334.58/11.47

Nw= 552.274N

Since Nw = Ff

Then Ff = 552.274N

Then, using frictional law

Ff = μs•Ng

The Ng doesn't change

Then, 552.274 = 1300μs

μs = 552.274/1300

μs = 0.42

c. Now, we want to know the maximum distance of the firefighter if the coefficient of static friction is reduce by half

Then, μs = 0.42/2

μs = 0.21

The assume the firefighter is at L2 from the horizontal.

Then, the frictional force is

Ff = μsNg

Ff = 0.21 × 1300

Ff = 273N

Then, Nw = Ff = 273N

Taking moment about point A

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•L2 + (490Cos55)•7 = (273Sin55)•14

464.6•L2 + 1967.37 = 3130.8

464.6•L2 = 3130.8—1967.37

464.6L2 = 1163.43

L2 = 1163.43/464.6

L2 = 2.5m

The maximum distance before the ladder begin to slip is 2.5m.

3 0

3 years ago

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The part of the atom that accounts for electricity is the

hichkok12 [17]

<span>The part of the atom that accounts for electricity is the "Electron"

In short, Your Answer would be Option A

Hope this helps!</span>

4 0

3 years ago

Read 2 more answers

All motion will eventually stop due to the "loss" of energy to___.