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NISA [10]
3 years ago
5

A Carnot engine operates with a cold reservoir at a temperature of 474 K and a hot reservoir at a temperature of 628 K. What is

the net entropy change (in J/K) as it goes through a complete cycle? slader
Physics
1 answer:
Zigmanuir [339]3 years ago
5 0

Answer:

zero

Explanation:

A Carnot engine is an ideal engine in which the process is cyclic and reversible.

A Carnot engine is the engine whose efficiency is 100 % and the entropy in is equal to entropy out. That means the net change in entropy is zero.

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An open pipe is 1.42 m long.
Varvara68 [4.7K]

Answer:

f(3) = 362.32 Hz

Explanation:

the formula of frequency of open pipe is

f(n) = (n+1)v/2L

n = 0,1,2,3,... (the order)

0 for the first

1 for the second

v = speed of sound

L = the lenght of pipe

f(3) = (2+1) * 343 / 2 * 1,42

f(3) = 3 * 343 / 2.84

f(3) = 362.324 Hz (or you can write 362.32 Hz

8 0
3 years ago
What is transferred by a force moving an object through a distance?
Komok [63]
<span>When you apply force to move an object at a distance, you are applying work. And work is energy in transit. The answer is letter D. For example, you see a cart at a distance. You observe that it is not moving. You want to transfer it to your backyard. You apply force to the cart and observed that the cart is not at the same position as it was before. You are applying work to the cart by transferring your energy to it.</span>
7 0
3 years ago
Read 2 more answers
Explain the different types of energy in a working wind turbine
tatiyna

Answer:

A wind turbine converts the kinetic energy of the wind into mechanical power.  

The wind moves the turbine and the turbine produce energy.

8 0
3 years ago
The displacement in simple harmonic motion is a maximum when the 1. velocity is a maximum. 2. kinetic energy is a maximum. 3. ve
dlinn [17]

Answer:

3. velocity is zero.

Explanation:

The velocity of a simple harmonic motion is given by

v = \omega\sqrt{A^2-x^2}

Here, <em>ω</em> is the angular velocity, <em>A</em> is the amplitude (or maximum displacement from the equilibrium point) and <em>x</em> is the displacement at any time.

At maximum displacement, <em>x </em>=<em> A</em>.<em> </em>Then

v = \omega\sqrt{A^2-A^2} = 0

Therefore, at maximum displacement, velocity is 0.

Practically, this can be observed in a simple pendulum. As it approaches the maximum displacement, its velocity reduces. It becomes zero at this point and then reverses as the pendulum changes course. Then the velocity begins to increase. It becomes maximum at the equilibrium point but once past that, the velocity begins to reduce as it approaches the other amplitude.

For acceleration,

a = -\omega^2x

It follows that at maximum displacement, the acceleration is a maximum. The negative sign indicates that it is in an opposite direction to the displacement. Both kinetic energy (\frac{1}{2}mv^2) and linear momentum (mv) are proportional to velocity; they are therefore both zero at the maximum displacement.

5 0
3 years ago
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0
3 years ago
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