Answer:
It is a well known fact that the earth rotates around the sun in an inclined axis which is approximately 23 degree. The inclined nature of earth axis causes variation in the solar heat received at any place on the earth surface. The hemisphere facing the sun due to this axial tilt, gets higher sun energy as compared to the opposite side. The hemisphere which faces the sun will experience summer whereas the hemisphere away from sun will experience winter.
In each of the hemisphere the polar areas will receive higher radiation and longer daytime during the summer season. However it has been observed that there is difference in radiation received at different areas of earth surface and radiated. The tropical areas have lower reflectance and thus a large part of incoming solar radiation have been absorbed along the tropics. The poles though have longer daytime during summer and hence greater solar radiation but due to high reflectance radiate more energy. Thus the tropical areas have surplus energy as compared to deficit energy areas of poles. This difference in energy creates a heat imbalance.
This net heat difference between poles and equator gives rise to a global circulation system leading to flow of heat from the net energy excess areas to deficit areas. This circulation takes place through atmosphere as well as oceans and different process of climate viz. evaporation, transpiration, rainfall, wind, convection, oceanic circulations etc work as tools of this system
Answer:
Explanation:
Given that
R=8 ft
Width= 10 ft
We know that hydro statics force given as
F=ρ g A X
ρ is the density of fluid
A projected area on vertical plane
X is distance of center mass of projected plane from free surface of water.
Here
X=8/2 ⇒X=4 ft
A=8 x 10=80 
So now putting the values
F=ρ g A X
F=62.4(32.14)(80)(4)
Answer:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
Answer:
1200KJ
Explanation:
The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.
P (rotor-loss) = 3 x K.E
P = 3 x 300 = 900 KJ
After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;
KE = 300 KJ
Since it is in opposite direction, it will also add up to rotor loss
P ( rotor loss ) = 900 + 300 = 1200 KJ
