How much work is done to move 2.10 μC of charge from the negative terminal to the positive terminal of a 3.50 V battery?

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

bixtya [17]

Answer:

a. $P=18.61\ W$

b. $\eta_c=0.6976=69.76\%$

c. $\eta=0.4718=47.18\%$

Explanation:

Given:

temperature of the hotter reservoir, $T_H=1250\ K$

temperature of the colder reservoir, $T_C=378\ K$

heat absorbed by the engine, $Q_H=1.42\times 10^{5}\ J$

heat rejected to the cold reservoir, $Q_L=7.5\times 10^4\ J$

time duration of the energy transfer, $t=1\ hr=3600\ s$

Now the work done by the engine:

Using energy conservation,

$W=Q_H-Q_L$

$W=14.2\times 10^4-7.5\times 10^4$

$W=6.7\times 10^4\ J$

a.

Hence the power output:

$P=\frac{W}{t}$

$P=\frac{6.7\times 10^4}{3600}$

$P=18.61\ W$

b.

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{378}{1250}$

$\eta_c=0.6976=69.76\%$

c.

now actual efficiency:

$\eta=\frac{W}{Q_H}$

$\eta=\frac{6.7\times 10^4}{1.42\times 10^{5}}$

$\eta=0.4718=47.18\%$

6 0

4 years ago

What is the energy equivalent of an object with a mass of 1. 05 g? 3. 15 Ă— 105 J 3. 15 Ă— 108 J 9. 45 Ă— 1013 J 9. 45 Ă— 1016 J

Molodets [167]

Considering the equivalence between mass and energy given by the expression of Einstein's theory of relativity, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

The equivalence between mass and energy is given by the expression of Einstein's theory of relativity, where the energy of a body at rest (E) is equal to its mass (m) multiplied by the speed of light (c) squared:

E=m×c²

This indicates that an increase or decrease in energy in a system correspondingly increases or decreases its mass, and an increase or decrease in mass corresponds to an increase or decrease in energy.

In other words, a change in the amount of energy E, of an object is directly proportional to a change in its mass m.

In this case, you know:

m=1.05 kg
c= 3×10⁸ m/s

Replacing:

E= 1.05 kg× (3×10⁸ m/s)²

Solving:

E= 9.45×10¹⁶ J

Finally, the correct answer is the last option: the energy equivalent of an object with a mass of 1.05 kg is 9.45×10¹⁶ J.

Learn more:

brainly.com/question/9477556

5 0

2 years ago

A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov

Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

4 0

3 years ago

Read 2 more answers

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

3 years ago

A capacitor is made by taking two sheets of aluminum foil, each 0.022 mm thick and placing between them a sheet of paper which c

olasank [31]

The answer is 18 nf in this equation

8 0

3 years ago