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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.

You might be interested in
A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.
Anna11 [10]

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

B = \frac{\mu_{0}NI}{L}

\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns

Since each turn has length 2πr of wire, the total length is:

L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

4 0
3 years ago
An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is
Alik [6]
We use a fundamental kinematic equation as follows:

V = Vo + g*t. 
<span>Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = </span><span>time to reach max. height </span>

<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>

<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>

<span>d = Vo*t + 0.5g*t^2. </span>
<span>d = 10*1 + 5*1^2 = 15 m. <---- OPTION C</span>
8 0
3 years ago
2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg &amp; object B’s mass is 11kg. After the colli
ss7ja [257]

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

3 0
3 years ago
John had a stroke and is having difficulty sitting and walking. Which medical professional should he consult to improve his phys
slava [35]
John needs to see a physical therapist because he cannot walk very well.
5 0
3 years ago
Read 2 more answers
Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
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