Answer:
The air fraction to be removed is 0.11
Given:
Initial temperature, T = 
= 283 K
Pressure, P = 250 kPa
Finally its temperature increases, T' = 
= 318 K
Solution:
Using the ideal gas equation:
PV = mRT
where
P = Pressure
V = Volume
m = no. of moles of gas
R = Rydberg's Constant
T = Temperature
Now,
Considering the eqn at constant volume and pressure, we get:
mT = m'T'
Thus
(1)
Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:
From eqn (1):
Answer:
The correct option is B
Explanation:
In part A pressure is given as 
In Part B we can calculate the pressure using the formula
Where F is the force which in this case is weight which can be obtained as follow
W = Mg
Substituting 70kg for M and 
for g
A is the area with a value 
So substituting into the above equation
for part c
A 
F 
Therefore 
From this we see that the value of P for part B is the greatest
The answer to this is
<span>A mass on a spring will oscillate vertically when it is lifted and released. The gravitational potential energy increases from a minimum at the lowest point to a maximum at the highest point. ... The total mechanical energy at those points is the sum of the elastic and gravitational potential energies.</span>
Hope This Helps!
:D
The answer to that is D. If u think about all u have to do is multiply 5 times 2 = 10
