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Simora [160]
3 years ago
14

A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at

2 bars to a final volume of 0.12 m3 , while being slowly heated through the base. The change in internal change of the gas 0.25 kJ. The piston and cylinder walls are fabricated from heat-resistant material and the piston moves smoothly in the cylinder. For the gas as the system, evaluate work and heat transfer each in kJ (Neglect the potential energy change and kinetic energy change).
Physics
1 answer:
Gnom [1K]3 years ago
7 0

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

W=p(V_f -V_i)

where in this case we have:

p = 2 bar = 2\cdot 10^5 Pa is the pressure

V_i = 0.1 m^3 is the initial volume

V_f = 0.12 m^3 is the final volume

Substituting,

W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ

The 1st law of thermodynamics also states that

\Delta U = Q-W

where

\Delta U is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

\Delta U = +0.25 kJ

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ

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emmasim [6.3K]

Answer:

The unit you should use for work done and energy is the joule (J) which is indeed the same as the newton metre (N m).

There is another physical quantity which is the product of force and distance and that is torque or moment of a force.

The unit you should use for torque is the newton metre (Nm) and not the joule.

Naming the units of work done and torque differently helps to emphasis the fact that work done and torque refer to two different physical quantities although the definitions of both quantities have the product of force and distance in them.

work done=force→⋅displacement→ and torque→=force→×displacement→

Hope I helped

4 0
3 years ago
A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f
zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

8 0
3 years ago
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A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10
irina [24]
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
5 0
3 years ago
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According to newtons law of action reaction what is the most likely to occur if to ice skaters with approximately the same mass
Free_Kalibri [48]

Answer:

They would keep on moving but unless being acted upon or stop slowly because of the friction

Explanation:

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3 years ago
A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i
Hoochie [10]

The acceleration of the box is approximately 1 m/s^2

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

\sum F = F_a - F_f

where

F_a = 20 N is the applied forward force

F_f=9.0 N is the friction force

Combining the two equations,

F_a-F_f=ma

And solving for the acceleration,

a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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