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Consider a sample of helium gas in a container fitted with a piston, as pictured below. The piston is frictionless, but has a ma

andrezito [222]

Question:

Consider a sample of helium gas in a container fitted with a piston as pictured below. The piston is frictionless, but has a mass of 10.0 kg. How many of the following processes will cause the piston to move away from the base and decrease the pressure of the gas? Assume ideal behavior.

I. Heating the helium. II.

II. toRemoving some of the helium from the container.

III. Turning the container on its side.

IV. Decreasing the pressure outside the container.

a) 0

b) 1

c) 2

d) 3

e) 4

Answer:

Only one process will cause the piston to move which is

i) Heating the helium

Explanation:

When helium is heated it becomes less dense or lighter. Heating the helium will cause an increase in volume which will make the piston to move away from the base. When the volume finishes increasing, the piston will stop moving which in turn will make the forces on both sides of the piston balanced, so the pressure inside will balance the weight of the piston and that of the atmosphere. From that we can see that there has been a pressure change as a result of heating.

6 0

3 years ago

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You have landed on an unknown planet, newtonia, and want to know what objects will weigh there. you find that when a certain too

AleksandrR [38]

Refer to the diagram shown.

Because the surface is frictionless, the resistive for, R, is zero.

Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²

The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg

For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²

Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)

3 0

3 years ago

Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the

svetoff [14.1K]

Answer:

$\mu_k=0.18$

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

$x: T+F-f_k=0\\\\y:N-mg=0$

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since $f_k=\mu_k N$ and $N=mg$ , we can rewrite the first equation as: