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3 years ago

What is the maximum efficiency of a heat engine whose operating temperatures are 590 ?c and 325 ?c?

1 answer:

7 0

<span>The maximum possible efficiency, i.e the efficiency of a Carnot engine , is give by the ratio of the absolute temperatures of hot and cold reservoir.
η_max = 1 - (T_c/T_h)

For this engine:
η_max = 1 - [ (20 +273)K/(600 + 273)K ] = 0.66 = 66%

The actual efficiency of the engine is 30%, i.e.
η = 0.3 ∙ 0.664 = 0.20 = 20 %

On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir:
η = W/Q_h

So the heat required from hot reservoir is:
Q_h = W/η = 1000J / 0.20 = 5000J</span>

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The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to

DedPeter [7]

Answer:

$9.4\cdot 10^{10} m$

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

$\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}$

where

$r_o$ is the distance of the new object from the sun (orbital radius)

$T_o=180 d$ is the orbital period of the object

$r_e = 1.50\cdot 10^{11} m$ is the orbital radius of the Earth

$T_e=365 d$ is the orbital period the Earth

Solving the equation for $r_o$ , we find

$r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m$

3 0

3 years ago

While refrigerant 410a is a near azeotropic refrigerant, it is still best when charging to remove the r-410a as a _______ from t

andreev551 [17]

While refrigerant 410a is a near azeotropic refrigerant, it is still best when charging to remove the r-410a as a liquid from the storage cylinder.
Azeotrope means a constant boiling mixture. it is a mixture of two or more liquids,by simple distillation whose proportions cannot be changed. A mixture behaving purely is azeotropic and the mixture which behave differently is called non-azeotropic.

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3 years ago

a wire with mass per unit length 75 g/m runs horizontally at right angles to a uniform horizontal 0.12 T magnetic field. what am

Sindrei [870]

Answer:

The amount of current that must flow through the wire for it to be suspended against gravity by magnetic force = 6.125 A

Explanation:

Force on a wire carrying current in an electric field is given by

F = (B)(I)(L) sin θ

For this question,

The magnetic force must match the weight of the wire.

F = mg

mg = (B)(I)(L) sin θ

(m/L)g = (B)(I) sin θ

Mass per unit length = 75 g/m = 0.075 kg/m

B = magnetic field = 0.12 T

I = ?

g = acceleration due to gravity = 9.8 m/s

θ = angle between wire's current direction and magnetic field = 90°

0.075 × 9.8 = 0.12 × I sin 90°

I = 0.075 × 9.8/0.12 = 6.125 A

3 0

3 years ago

Pls pls help me out AHH, what is not true about MEIOSIS?

agasfer [191]

Answer:

Explanation:

This is not true as the number of chromosomes in the daughter cells are half the number in the parent's cells

4 0

3 years ago

1. A 4000-kg truck traveling with a velocity of 20 m/s due south collides headon with a 1350-kg car traveling with a velocity of

velikii [3]

(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) The size of momentum is 93500 kgm/s and it will be directed towards South.

Explanation:

The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.

(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.

$Momentum of truck = Mass * Speed = 4000 * 20 =80000 kgm/s$

Similarly, the momentum of car will be

$Momentum of car = 1350 * 10 = 13500 kgm/s$

So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.

Momentum of vehicles after collision = total momentum before collision

Momentum after collision = 80000+13500 = 93500 kgm/s.

The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.

Thus, the size of momentum is 93500 kgm/s and it will be directed towards South.

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2 years ago