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
Ozone in troposphere is also know as Bad Ozone, Evil Ozone and Ground Level Ozone.
The position of the first ball is
while the position of the second ball, thrown with initial velocity 
, is
The time it takes for the first ball to reach the halfway point satisfies
We want the second ball to reach the same height at the same time, so that
A net torque of magnitude is 600
Answer:
The magnitude of the angular acceleration is α = (3 * F)/(M * L)
Explanation:
using the equation of torque to the bar on the pivot, we have to:
τ = I * α, where
I = moment of inertia
α = angular acceleration
τ = torque
The moment of inertia is equal to:
I = (M * L^2)/3
Also torque is equal to:
τ = F * L
Replacing:
I * α = F * L
α = (F * L)/I = (F * L)/((M * L^2)/3) = (3 * F)/(M * L)
