To get the solution you must need to draw a force triangle. Attach the head of the 60N north force arrow with the tail of the 60N east force arrow. The subsequent is the arrow connecting he tail and head of the two arrows.
You get a right angled triangle, and the resultant is (60^2 + 60^2) ^0.5 = 84.85 N or 85 N northeast.
Answer:
Δe=0.578 kJ/kg
Explanation:
Given data
Velocity v₁=0 m/s
Velocity v₂=34 m/s
to find
Specific energy change Δe
Solution
The specific energy change is simply determined from change in velocity
Δe=(v₂²-v₁²)/2
Put the given values to find the specific energy change

Δe=0.578 kJ/kg
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):


<span>B. Energy is never created nor destroyed.
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