Answer:
Relative to the ground, the velocity of the aircraft is 240 km/hr
Explanation:
Relative velocity is different from normal velocity;
When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;
This is known as the relative velocity;
The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;
The wind is in the opposite direction at 80 km/hr;
R = relative velocity of the aircraft
v = actual velocity of the aircraft
w = velocity of the wind
R = v + w
Note: if the wind was moving in the same direction, the formula would be R = v - w
320 = v + 80
v = 320 - 80
v = 240
The velocity relative to the ground is simply the actual velocity as the ground doesn't move;
So, relative to the ground, the velocity of the aircraft is simply 240 km/hr
<u>Answer</u>
The combined displacement is 2km north
<u>Explanation</u>
Since displacement is a vector quantity, we take into account the direction.
Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.
We now add to obtain,
This will simplify to
Therefore the combined displacement is 2km north
300N/25 kg= divide them for the answer
Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁
