In general terms, the efficiency of a system can be thought of as the output per unit input. Which of the expressions is a good mathematical representation of efficiency e of any heat engine?
Where,
Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem
Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem
W: the amount of work done by the engine during one cycle or during some time specified in the problem

A) e=QhW
B) e=QcQh
C) e=QcW
D) e=WQh
E) e=WQc

Answers

Answer 1

Answer:

Answer:

Efficiency e = W/Qh

Explanation:

As written above efficiency of a system is calculated as the output per unit input. For heat Engine, Efficiency is calculated by dividing the Work done by Engine by Heat absorbed from hot reservoir.

In theoretical terms The maximum efficiency of a heat engine (which no engine ever attains) is equal to the temperature difference between the hot and cold ends divided by the temperature at the hot end, each expressed in absolute temperature (Kelvin).

But in practical calculations, it is calculated as e = W/Qh , and we define the thermal efficiency, of any heat engine as the ratio of the work it does, W, to the heat input at the high temperature, Qh.

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Draw a ray diagram for an object placed more than two focal lengths in front of a converging lens.

Answers

Answer:

Explanation:

There is a convex lens M N is placed. An object AB is placed at a distance more than two focal lengths of the lens.

A ray of light is starting from point A and parallel to the principal axis, then after refraction it goes from the focus.

Another ray which goes through the optical centre of the lens becomes undeviated after refraction.

The two refracted rays meet at the point A', So A'B is the image of AB.

The nature of image is real, inverted and diminished.

(a) A woman climbing the Washington Monument metabolizes 6.00×102kJ of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?

Answers

Answer:

492 kJ

Consistent

Explanation:

Q = Heat stored by woman from food = 600 k J

η = Efficiency of woman = 18% = 0.18

Q' = heat transferred to the environment

heat transferred to the environment is given as

Q' = (1 - η) Q

Inserting the values

Q' = (1 - 0.18) (600)

Q' = 492 kJ

Yes the amount of heat transfer is consistent. The process of sweating produces the heat and keeps the body warm

Final answer:

A woman climbing the Washington Monument metabolizes food energy with 18% efficiency, meaning 82% of the energy is lost as heat. When we calculate this value, we find that 492 kJ of energy is released as heat, which is consistent with the fact that people quickly warm up when exercising.

Explanation:

The woman climbing the Washington Monument metabolizes 6.00×10² kJ of food energy with an efficiency of 18%. This implies that only 18% of the energy consumed is used for performing work, while the remaining (82%) is lost as heat to the environment.

To calculate the energy lost as heat:

Determine the total energy metabolized, which is 6.00 × 10² kJ.
Multiply this total energy by the percentage of energy lost as heat (100% - efficiency), which gives: (6.00 × 10² kJ) * (100% - 18%) = 492 kJ.

The released heat of 492 kJ is consistent with the fact that a person quickly warms up when exercising, because a significant portion of the body's metabolic energy is lost as heat due to inefficiencies in converting energy from food into work.

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A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V. Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Answers

Answer:

U_eq = 1.99 * 10^(-10) J

Explanation:

Given:

Plate Area = 10 cm^2

d = 0.01 m

k_dielectric = 3

k_air = 1

V = 15 V

e_o = 8.85 * 10 ^-12 C^2 / N .m

Equations used:

U = 0.5 C*V^2 .... Eq 1

C = e_o * k*A /d .... Eq 2

U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3

For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:

U_electric = 0.5 e_o * k_1*A*V^2 /2*d

U_air = 0.5 e_o * k_2*A*V^2 /2*d

The total Energy is:

U_eq = U_electric + U_air

U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d

U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d

Plug the given values:

U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01

U_eq = 1.99 * 10^(-10) J

The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

v = 4.1 m / s

Explanation:

Velocity is defined by the relation

v = $(dx)/(dt)$

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

v = $(\Delta x)/(\Delta t)$

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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The pressure in a section of horizontal pipe with a diameter of 2.5 cm is 139 kPa. Water ï¬ows through the pipe at 2.9 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should be the diameter of the constricted section? The acceleration of gravity is 9.81 m/s2 . Assume laminar nonviscous ï¬ow.

Answers

Answer:

d = 2*0.87 = 1.75 cm

Explanation:

by using flow rate equation to determine the speed in larger pipe

$\phi =\pi r^2 v$

$v = (\phi)/(\pi r^2)$

$= (2900 cm^3/s)/(3.14(1.25cm)^2)$

= 591.10 cm/s

= 5.91 m/s

by Bernoulli's EQUATION

$p1 +(1)/(2) \rho v1^2 = p2 +(1)/(2) \rho v2^2$

$139000+ (1)/(2)*1000*5.91^2 = 101000 +(1)/(2)*1000* v2^2$

solving for v2

v2 = 10.53 m/s

diameter can be determine by using flow rate equation

$q = v \pi r^2$

$r^2 = (q)/(\pi v)$

$= (2900)/(3.14*1053)$

r = 0.87 cm

d = 2*0.87 = 1.75 cm

Two charged metallic spheres of radii, R = 10 cms and R2 = 20 cms are touching each other. If the charge on each sphere is +100 nC, what is the electric potential energy between the two charged spheres?