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Readme [11.4K]
2 years ago
14

Two blocks with different temperatures had entropies of 10 J/K and 30 J/K before they were brought in contact. What can you say

about the entropy of the combined system after the two came in contact with each other?a.Ssystem > 40 J/Kb.Ssystem = 20 J/K c.Ssystem < 20 J/Kd.Ssystem > 20 J/Ke.Ssystem = 40 J/Kf.Ssystem < 40 J/K
Physics
1 answer:
True [87]2 years ago
6 0

Answer:

a. Ssystem  > 40 J/K

Explanation:

Given that

The entropy of first block = 10 J/K

The entropy of second block = 30 J/K

When two bodies come into contact with each other, the entropy of the combined system will increase and the entropy sum remains unchanged: According to the Second law of thermodynamics.The entropy of the system will be greater than 40  J/K.

Therefore the answer is a.

Ssystem  > 40 J/K

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A mass m is placed at the rim of a frictionless
olchik [2.2K]

Answer:

Explanation:

at collision, m has a velocity of √2gR

The center of mass of the system has a velocity of

m(√2gR) + 3m(0) = (m + 3m)v

v = ¼√2gR

The stationary mass (3m) will depart at twice the system center of mass velocity

v(3m) = ½√2gR   ANSWER B

The relative velocity before the collision will be the negative of the relative velocity after the collision.

v(m) = ½√2gR - √2gR = -½√2gh ANSWER A

m will rise to a height of v²/2g = (-½√2gR)²/2g = 0.25R

7 0
2 years ago
A marble is thrown horizontally with a speed of 7 m/s. When the marble lands, it hastraveled a horizontal displacement of 30 m.
Alex Ar [27]

Answer:

t = 4.28 s

Explanation:

The horizontal speed of a marble = 7 m/s

The horizontal displacement covered by the marble = 30 m

We need to find the time for which the marble is in air. Let the time is t. Using the formula for speed to find it as follows :

s=\dfrac{d}{t}\\\\t=\dfrac{d}{s}\\\\t=\dfrac{30\ m}{7\ m/s}\\\\=4.28\ s

So, it will in the air for 4.28 s.

6 0
3 years ago
The top of a ladder slides down a vertical wall at a rate of 0.125 m/s. at the moment when the bottom of the ladder is 5 m from
Contact [7]
Vertically (y-axis), horizontally(x-axis)

dy/dt = -0.125 m/s (-ve since y decreasing )

dx/dt = +0.3 m/s (+ve, x increases)

and, x = 5 m

length of the ladder = k (a constant)

k^2 = x^2 + y^2

differentiating it wrt t,

0 = 2x dx/dt + 2y dy/dt

0 = 2(5)(0.3) + 2(y)(-0.125)

y = 12

which means, when the bottom of the ladder is 5m from the wall, the top of the ladder is 12m from the bottom.

thus, k^2 = 5^2 + 12^2

length of the ladder, k = 13 m
5 0
3 years ago
A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the
Vitek1552 [10]

Answer:

147 N

Explanation:

static friction = frictional force max / normal

normal force = mg = 50 kg × 9.8 m/s² = 490 N

0.3 = Frictional force / 490 N

0.3 × 490 N = 147 N

The body will not move because the frictional force is greater than the force applied.

4 0
3 years ago
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Which of the following is an example of qualitative data
NemiM [27]
Qualitative data is easy to explain. You either agree or disagree in writing. Wrong. You agree, disagree, or qualify. Meaning agree to somewhat extend, but not entirely. So you can agree that people should walk instead of talk in the progressive manner, but not in the impulsive manner.
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3 years ago
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