Answer:
a) 75.5 degree relative to the North in north-west direction
b) 309.84 km/h
Explanation:
a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it
So the pilot should head to the West-North direction at an angle of
relative to the North-bound.
b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is
Answer:
Temperature : 92.9 F
Internal Energy change: -2.53 Btu/lbm
Explanation:
As
mh1=mh2
h1=h2
In table A-11 through 13E
p2=120Psi, h1= 41.79 Btu/lbm,
u1=41.49
So T1=90.49 F
P2=20Psi
h2=h1= 41.79 Btu/lbm
T2= -2.43F
u2= 38.96 Btu/lbm
T2-T1 = 92.9 F
u2-u1 = -2.53 Btu/lbm
Explanation:
please send full question....
Answer:
Explanation:
A and B are in series , Total resistance = Ra + Rb
This resistance is in parallel with single resistor C
Equivalent resistance Re = Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )]
Now this combination is in series in single resistance D .
Total resistance = Rd + Re
= Rd + { Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )] }
As per the question, the mass of meteorite [ m]= 50 kg
The velocity of the meteorite [v] = 1000 m/s
When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.
We are asked to calculate the gain in kinetic energy of earth.
The kinetic energy of meteorite is calculated as -
![Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2](https://tex.z-dn.net/?f=Kinetic%5C%20energy%5C%20%5BK.E%5D%5C%20%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
![=\frac{1}{2}50kg*[1000\ m/s]^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D50kg%2A%5B1000%5C%20m%2Fs%5D%5E2)
Here, J stands for Joule which is the S.I unit of energy.
