Answer:
The work flow required by the compressor = 100.67Kj/kg
Explanation:
The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .
The work flow can be determined using the equation:
M1h1 + W = Mh2
U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2
Workflow = P2alpha2 - P1alpha1
Workflow = (h2 -U2) - (h1 - U1)
Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)
Workflow = ( 193.191 - 92.519)Kj/kg
Workflow = 100.672Kj/kg
Answer:
You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.
Explanation:
spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.
Let any rolling ball of mass (m ) is traveling with velocity v ,
common effect on ball (N) = mg
because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.
hence,
fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.
-17.555m/s
first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.
then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.
then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2
Answer:
Kf= 36 J
W(net) = 32 J
Explanation:
Given that
m = 2 kg
F= 4 N
t= 2 s
Initial velocity ,u= 2 m/s
We know that rate of change of linear momentum is called force.
F= dP/dt
F.t = ΔP
ΔP = Pf - Pi
ΔP = m v - m u
v= Final velocity
By putting the values
4 x 2 = 2 ( v - 2)
8 = 2 ( v - 2)
4 = v - 2
v= 6 m/s
The final kinetic energy Kf
Kf= 1/2 m v²
Kf= 0.5 x 2 x 6²
Kf= 36 J
Initial kinetic energy Ki
Ki = 1/2 m u²
Ki= 0.5 x 2 x 2²
Ki = 4 J
We know that net work is equal to the change in kinetic energy
W(net) = Kf - Ki
W(net) = 36 - 4
W(net) = 32 J