<span>The change in internal energy is only gravitional PE because the tube is being drug up at a constant speed. Since it is at a constant speed, the change in KE is 0.
Change in PE = m*g*h = 78 kg * 10 m/s^2 * 30 m = 23400 J
Work done on the system is from the force
Work = force * distance = 350 N * 120 m = 42000 J
So, work added 42000 J to the system, but the rider's energy only increased 23400 J. Therefore, friction took up the difference. Friction is where the thermal energy comes from
Q = 42000 J - 23400 J = 18600 J.
Therfore, friction generated 18600 J of heat to the surroundings.</span>
According to Ohm's Law, the resistance, current and voltage are related as:
V = IR
⇒
R = V/I
V is given to be 12 Volts.
I is given to be 0.5 Ampere
So, resistance will be:
R = 12/0.5 = 24 ohms
Answer:
option (b)
Explanation:
Let the resistance of each resistor is R.
In series combination,
The effective resistance is Rs.
rs = r + R + R + .... + n times = NR
Let V be the source of potential difference.
Power in series
Ps = v^2 / Rs = V^2 / NR ..... (1)
In parallel combination
the effective resistance is Rp
1 / Rp = 1 / R + 1 / R + .... + N times
1 / Rp = N / R
Rp = R / N
Power is parallel
Rp = v^2 / Rp = N V^2 / R ..... (2)
Divide equation (1) by equation (2) we get
Ps / Pp = 1 / N^2
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
