Just need some help trying to figure out some of these practice questions. Please and thank you!!!
1 answer:
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Answer:
(a) 83475 MW
(b) 85.8 %
Explanation:
Output power = 716 MW = 716 x 10^6 W
Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3
mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000
= 1.35 x 10^8 kg
Time, t = 1 hr = 3600 second
T1 = 25.4° C, T2 = 30.7° C
Specific heat of water, c = 4200 J/kg°C
(a) Total energy, Q = m x c x ΔT
Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J
Power = Energy / time
Power input =
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
Thus, the efficiency is 85.8 %.
Answer:
A) 667 J
B) 381.4 J
C) 0 J
D) 245.4 J
E) 40.2J
F) 2 m/s
Explanation:
Let g = 9.81 m/s2
A) The work done on the suitcase is the product of the force applied and the distance travelled:
w = Fs = 145 * 4.6 = 667 J
B) The work done by gravitational force the dot product between the gravity vector and the distance vector
C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0
D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient
E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work
F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that: