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.
0.74in
Explanation:
Given parameters:
length range of adult mosquito = 3.0 - 6.0mm
Smallest mosquito = 2.5mm
Largest mosquito = 19mm
Average mass range = 3.0 - 5.0mg
Unknown:
Length of largest known mosquito in inches = ?
Solution;
The length is the longest dimension. It is how long a body is.
The problem here is converting from mm to inches;
The length of the longest mosquito which is the largest is 19mm
19mm to inches;
1mm = 0.039inches
19mm = 19mm x 
= 0.74in
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Answer:
Cells are small because they need to keep a surface area to volume ratio that allows for adequate intake of nutrients while being able to excrete the cells waste.
Explanation:
That is why the cell needs to be small
Answer:
80 m/s
Explanation:
Given:
a = -5 m/s²
v = 0 m/s
Δx = 640 m
Find: v₀
v² = v₀² + 2a(x − x₀)
(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)
v₀ = 80 m/s
Answer:570.54 N
Explanation:
Given
mass of man=76 kg
As man is standing over inclined building therefore
its weight has two components i.e. sin and cos component
Force perpendicular to inclined wall
F=570.54 N
