There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.
I think this is correct, but I am not entirely certain.
Find the force constant of the spring:
F = - KX
(0 - 62.4) = -K(0.172m)
-362.791 = -K
362.791 N/m = K
Find the work done in stretching the spring:
W = (1/2)KX
W = (1/2)(362.791)(0.172m)
W = 31.2 J
Answer:
a . 0.35cm
b. 11.33cm
Explanation:
a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:
Hence, for currents in same direction, the point is 0.35cm
b. Given both currents flow in opposite directions, the null point lies on the other side.
#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:
Let x be distance of N from first wire, then distance from 2nd wire is 4+x:
Hence, if currents are in opposite directions the point on x-axis is 11.33cm
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:
c) Both the parts will weigh the same
Explanation:
center of gravity is based on weight so if you cut down the center of gravity you would have 2 equal parts
(might be D if it is cutting against the center of gravity)
