A sample of an ideal gas at Pi is initially confined to one chamber of the apparatus represented above, and the other chamber is
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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
(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')
limit is 0 → 4√3
On solving the equation, we get the value of limit as 32√3
Therefore, the force on one end of the trough is 5.4 X 10⁵ N
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
The magnitude of the average friction force acting on the toboggan while it was moving is 10.28571 N
Answer:264.58 mm of Hg
Explanation:
Given
Volume of gas
Pressure
If the Plunger is pressed volume is reduced to
As temperature of the gas does not change so it follows boyle's law
According to which