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4 years ago

A heat engine is a device that uses _____ to produce useful work.

Physics

2 answers:

Troyanec [42]4 years ago

8 0

Heat engine is a device which is used to convert the heat energy into useful work

Heat engine extracts heat from high temperature reservoir and then convert that extracted heat into useful work

this process efficiency is calculated by the formula

$\eta = 1 - \frac{T_2}{T_1}$

here we know that

$T_1$ = temperature of high temperature reservoir

$T_2$ = temperature of low temperature reservoir

so now from this we can say that high efficiency heat engine will convert more work

so here correct answer will be

HEAT

Rainbow [258]4 years ago

5 0

Does it have have a word bank if not then gasoline

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A heat engine operates with 70.0 kcal of heat supplied and exhausts 30.0 kcal of heat. How much work did the engine do?

gladu [14]

The work done by the heat engine is 40 kCal.

The given parameters;

input heat of the engine, Q₁ = 70 kCal

output heat of the engine, Q₂ = 30 kCal

To find:

the work done by the heat engine

The work done by the heat engine is the change in the heat energy of the engine;

W = Q₂ - Q₁

Substitute the given parameters and solve work done (W)

W = 70 kCal - 30 kal

W = 40 kCal

Thus, the work done by the heat engine is 40 kCal.

Learn more here: brainly.com/question/4280097

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3 years ago

Which situation is an an example of competition that could be found in the grassland

astraxan [27]

A good answer is a giraffe and a tree

4 0

3 years ago

Read 2 more answers

Two long, parallel wires are separated by 2.0 m. Each wire has a 28-A current, but the currents are in opposite directions. Part

BlackZzzverrR [31]

Given Information:

Current = I = 28 A

distance between wires = r = 2.0 m

Required Information:

Magnetic field = B = ?

Answer:

B = 12x10⁻⁶ T

Step-by-step explanation:

Biot-Savart Law is given by

B = μ₀I/2πr

Where μ₀ is the permeability of free space, I is the current flowing through the wire and B is the magnitude of the magnetic field produced.

We are asked to find the magnetic field midway between the wires so r/2 = 1

B = 4πx10⁻⁷*28/2π*1

B = 6x10⁻⁶ T

since the same amount of current flows in both wires therefore, equal amount of magnetic field will be produced in both wires

B = 2*6x10⁻⁶ T

B = 12x10⁻⁶ T

Therefore, the net magnetic field midway between the two wires is 12x10⁻⁶ T.

3 0

3 years ago

A particle with charge q and momentum p, initially moving along the x-axis, enters a region where a uniform magnetic field* B=(B

VMariaS [17]

Answer:

Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

Explanation:

So, from the question, we are given that the charge = q, the momentum = p.

=> From the question We are also given that, "initially, there is movement along the x-axis which then enters a region where a uniform magnetic field* B = (B0)(k) which then extends over a width x = L, the distance = d in the +y direction as it traverses the field."

Momentum,P = mass × Velocity, v -----(1).

We know that for a free particle the magnetic field is equal to the centrepetal force. Thus, we have the magnetic field = mass,.m × (velocity,v)^2 / radius, r.

Radius,r = P × v / B0 -----------------------------(2).

Centrepetal force = q × B0 × v. ----------(3).

(If X = L and distance = d)Therefore, the radius after solving binomially, radius = (d^2 + 2 L^2) / 2d.

Equating Equation (2) and (3) gives;

P = B0 × q × r.

Hence, the Magnitude of momentum = q × B0 × [d^2 + 2L^2] / 2d.

4 0

3 years ago

What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 1

-BARSIC- [3]

Answer:

404K

Explanation:

Data given, Kinetic Energy.K.E=8.37*10^-21J

Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.

This relationship can be expressed as

$K.E=\frac{3}{2}KT$

where K is a constant of value 1.38*10^-23

Hence if we substitute the values, we arrive at

$T=\frac{2/3(8.37*10^{-21})}{1.38*10^-23}\\ T=404K$

converting to degree we have $131^{0}C$

4 0

3 years ago