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:
1.2727 stokes
Explanation:
specific gravity of fluid A = 1.65
Dynamic viscosity = 210 centipoise
<u>Calculate the kinematic viscosity of Fluid A </u>
First step : determine the density of fluid A
Pa = Pw * Specific gravity = 1000 * 1.65 = 1650 kg/m^3
next : convert dynamic viscosity to kg/m-s
210 centipoise = 0.21 kg/m-s
Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A
= 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec
Convert to stokes = 1.2727 stokes
Answer and Explanation:
<u>The correct answer choice is Contacting Overhead Power Lines.</u>
This is because the powerlines are making contact (touching) his car, and the example isn't related to the other answer choices.
<u><em>#teamtrees #PAW (Plant And Water)</em></u>
Answer:
Evaporation.
Explanation:
Evaporation is the stage of the Water Cycle where water turns into water vapor. The steps following Evaporation in order include Condensation, Precipitation, and Transpiration.
Answer:
Velocity of ball B after impact is
and ball A is 
Explanation:
= Initial velocity of ball A

= Initial velocity of ball B = 0
= Final velocity of ball A
= Final velocity of ball B
= Coefficient of restitution = 0.8
From the conservation of momentum along the normal we have

Coefficient of restitution is given by



Adding the above two equations we get



From the conservation of momentum along the plane of contact we have


Velocity of ball B after impact is
and ball A is
.