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BartSMP [9]
2 years ago
9

A sample of an ideal gas at Pi is initially confined to one chamber of the apparatus represented above, and the other chamber is

initially evacuated. The valve connecting the chambers is opened, and the gas expands at constant temperature to fill both chambers. What best describes ΔS for the process?
Physics
1 answer:
stiv31 [10]2 years ago
7 0

If the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, its temperature does not change.

When the valve of the chamber opens, the ideal gas expands and since, it is free to move its entropy increases. Entropy is a measure of uncertainty or randomness. Thus, ΔS becomes positive.

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Show by algebraic reasoning that your gravitational acceleration toward an object of mass M a distance d away is a = GM/d2 and t
trasher [3.6K]
Gravitational force is 
GmM / r^2

Force is also m x a

When equating both of them, m cancels out, which leaves us with GM/r^2

Hope this helps
8 0
2 years ago
A trough is filled with a liquid of density 875 kg/m3. The ends of the trough are equilateral triangles with sides 6 m long and
k0ka [10]

The force on one end of the trough is 5.4 X 10⁵ N

<u>Explanation:</u>

The triangle is equilateral which means all the interior angles are 60° and the sides are 6m long.

According to the figure,

AI / 8 = sin (60) = √3/2

AI = 4√3

The depth of the water is AI = 4√3

The interval becomes, | 0 , 4√3|

w = 2JK

w = \frac{2}{\sqrt{3} } (4\sqrt{3} - y')\\\\P_i = pgy\\\\F_i = pgy' X \frac{2}{\sqrt{3} } (4\sqrt{3} - y')dy                         (the hydrostatic force acting on the strip is the product of the pressure and the area)

where.

ρ = 875 kg/m³

g = 9.8m/s²

d = depth ( d = y')

\lim_{n \to \infty} E^n_1 F_i\\\\ \lim_{n \to \infty} E^n_1 pgy' \frac{2}{\sqrt{3} } (4\sqrt{3}  - y')dy\\\\= \int\limits^4_0 {pgy \frac{2}{\sqrt{3} }(4\sqrt{3} - y)  } \, dy\\\\= \frac{2pg}{\sqrt{3} } \int\limits^4_0 {4\sqrt{3} - y) } \, dy

limit is 0 → 4√3

On solving the equation, we get the value of limit as 32√3

F = \frac{2pg}{\sqrt{3} } X 32\sqrt{3}  = 64pg\\\\F = 64 X 875 X 9.8\\\\F = 548800N

Therefore, the force on one end of the trough is 5.4 X 10⁵ N

3 0
2 years ago
At one particular moment, a 15.0 kg toboggan is moving over a horizontal surface of snow at 4.80 m/s. After 7.00 s have elapsed,
GuDViN [60]

Answer:

10.28571 N

Explanation:

m = Mass of toboggan = 15 kg

u = Initial velocity = 4.8 m/s

v = Final velocity = 0

t = Time taken = 7 seconds

Friction force is given by the change in momentum over time

F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{15(0-4.8)}{7}\\\Rightarrow F=-10.28571\ N

The magnitude of the average friction force acting on the toboggan while it was moving is 10.28571 N

3 0
1 year ago
What may be a serious side effect of anti-depressants specific in adolescents?
Klio2033 [76]

I think its suicidal ideation......

I think

3 0
2 years ago
A gas is placed in a 5.0-mL syringe. It exerts 127 mm Hg of pressure on the inside walls of the syringe. The syringe's plunger i
Gekata [30.6K]

Answer:264.58 mm of Hg

Explanation:

Given

Volume of gas V_1=5\ mL

Pressure P_1=127\ \text{mm of Hg}

If the Plunger is pressed volume is reduced to

V_2=2.4\ mL

As temperature of the gas does not change so it follows boyle's law

According to which

P_1V_1=P_2V_2

127\times 5=P_2\times 2.4

P_2=264.58\ \text{mm of Hg}

8 0
1 year ago
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