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Readme [11.4K]
3 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]3 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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The height of the tide measured at a seaside community varies according to the number of hours t after midnight. If the height h
antoniya [11.8K]

Explanation:

Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :

h=-\dfrac{1}{2}t^2+6t-9

When the tide first be at 6 ft, put h = 6 ft in above equation as :

-\dfrac{1}{2}t^2+6t-9=6

-t^2+12t-18=0

On solving the above equation to find the value of t. It is equal to :

t = 3.551 seconds

or

t = 8.449 seconds

So, the tide of 6 ft is at  3.551 seconds and 8.449 seconds. Hence, this is the required solution.

6 0
3 years ago
A block has two strings attached to it on opposite ends. One string has a force of 5 N,
juin [17]

Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

• one rope pulling in one direction with magnitude 15 N,

• the other rope pulling in the opposite direction with mag. 5 N, and

• friction, opposing the direction of the block's motion, with mag. 3 N.

It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = <em>m</em> (1 m/s²)

where <em>m</em> is the mass of the block. Solve for <em>m</em> :

7 N = <em>m</em> (1 m/s²)

<em>m</em> = (7 N) / (1 m/s²)

<em>m</em> = 7 kg

(B) The friction force is proportional to the normal force, so that if <em>f</em> is the mag. of friction and <em>n</em> is the mag. of the normal force, then <em>f</em> = <em>µ</em> <em>n</em> where <em>µ</em> is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. <em>w</em>) cancel out:

<em>n</em> + (-<em>w</em>) = 0

<em>n</em> = <em>w</em>

<em>n</em> = <em>m</em> <em>g</em>

where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity.

<em>n</em> = (7 kg) (9.8 m/s²)

<em>n</em> = 68.6 N

Then

3 N = <em>µ</em> (68.6 N)

<em>µ</em> = (3 N) / (68.6 N)

<em>µ</em> ≈ 0.044

4 0
3 years ago
The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grainsare there in the ball
Alina [70]

Answer:

the number of grains in the ball is 274,848

Explanation:

Given that;

diameter = 0.5 mm

so radius r = 0.25 mm

first we determine the volume of the ball using the following equation;

V = 4/3×πr³

we substitute

V = 4/3×π(0.25)³

V =  0.06544 mm³

Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³

so;

Number of grains N = 0.06544 × 4,200,000

N = 274,848 grains

Therefore, the number of grains in the ball is 274,848

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Anarel [89]

Answer:

C. Bacteria

Explanation:

Bacteria and archaea are the two domains which consist of organisms with prokaryotic cells. Prokaryotic cells are cells without nuclei or membrane-bound organelles.

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Do u know the formula
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