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kogti [31]
3 years ago
9

A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in2 and weight of 100 lbf. The atmo

sphere exerts a pressure of 14.7 lbf/in2 on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases by 1 ft. The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the change in internal energy of the gas, in Btu.
Physics
1 answer:
kompoz [17]3 years ago
6 0

Answer:

ΔU=0.8834 Btu

Explanation:

Given data

Area of piston A=40 in²

The weight W=100 lbf

Atmospheric Pressure P=14.7 lbf/in²

Work added E=3 Btu

The change in elevation Δh=1 ft =12 inch

To find

Change in internal energy of the gas ΔU

Solution

For Piston

ΔPE=| W+(P×A)×Δh |

ΔPE=| 100+(14.7×40)×12 |

ΔPE=8256 lbf.in

ΔPE=8256×0.000107

ΔPE=0.8834 Btu

From law of conservation of energy then ,the charging in the potential energy of the piston is made by exerting force by gas

Wgas= -ΔPE

Wgas= -0.8834 Btu

For the gas as a system and by applying first law of thermodynamics

Q-W=ΔU

0-(-0.8834 Btu)=ΔU

So

ΔU=0.8834 Btu

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Answer:

Explanation:

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Angle with horizontal Ф

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A spherical shell with a net charge of 3Q surrounds a point charge of -q at the center of the shell. The charges on the inner an
aleksley [76]

Answer:

1) The charge on the outer shell is +4·Q

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Explanation:

1) The given parameters of the spherical shell are;

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The point charge surrounded by the spherical shell = -Q

Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;

The net charge, 3·Q = -q + x

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The charge on the outer shell, x = 4·Q

2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;

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A 3.00-kg rifle fires a 0.00500-kg bullet at a speed of 300 m/s. Which force is greater in magnitude:(i) the force that the rifl
alekssr [168]

Answer:

C. both forces have the same magnitude

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Here the action force is equal to the reaction force in accordance with the Newton's third law of motion.

Also when we apply the conservation of momentum so that the momentum bullet and the momentum of the gun are equal and according to the second law of motion by Newton, we have force equal to the rate of change in momentum.

We have the equation for momentum as:

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Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

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F= -495.65 × 10^-3/ 1.29 × 10^-3

F= -384.22N

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