Answer:
Explanation:
recall that power is energy carried out or work done per time
P=W/t
P=2*10^6*35
t=6*60=420S
W=Energy
E=2*10^6*35*360S
E=25200000000
Energy stored by water from rest is called potential energy. Since the water is falling from a height , we calculate potential energy as thus
E=M*g*h
Assume that the water intakes are effectively 175 m above the electric generators. How much water must pass through the generators to power 2 million 35-W Las Vegas light bulbs for 6.0 minutes?
M=mass of water
g=acceleration due to gravity 9.81m/s^2
h=height ,175m
25200000000=M*9.81*175
M=
M=1716.75kg
Answer:
We feel cold when tap or well water in winter because heat flows from hot body to cold body.
Explanation:
Our <em>body</em><em> </em><em>is</em><em> </em><em>in</em><em> </em><em>optimal</em><em> </em><em>status</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>hot</em><em> </em><em>body</em><em> </em><em>and</em><em> </em><em>tap</em><em> </em><em>or</em><em> </em><em>we</em><em>ll</em><em> </em><em>water</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>cold</em><em> </em><em>body</em><em>.</em><em> </em><em>Theref</em><em>ore</em><em> </em><em>we</em><em> </em><em>feel</em><em> </em><em>cold</em><em>.</em>
Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
Answer:
when a two bodies A and B are moving at an angle 180° with each other then the relative velocity is the sum of bodies the velocity .i.e,
when the bodies move the opposite direction
then their relative velocity is the sum of individual velocities.
Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.