Answer:
157.8 J
Explanation:
m = mass of the cylinder = 7 kg
h = height difference in top and bottom of the incline = 2.3 m
g = acceleration due to gravity = 9.8 m/s²
TE = Total Energy at the bottom
PE = Gravitational potential energy at the top
Using conservation of energy
Total Energy at the bottom = Gravitational potential energy at the top
TE = PE
TE = m g h
TE = (7) (9.8) (2.3)
TE = 157.8 J
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
<u>A) Determine how long the clock run on the battery. use the relation below</u>
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
<u>B) Determine how many electrons per second flowed </u>
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
Vf = Vo + at
Vf = 20 m/s
Vo = 50 m/s
a = ?
t = 15
Therefore
20 = 50 + 15a
20 - 50 = 15a
-30 = 15a
a = -30 / 15
a = -2 m/s²
Explanation:
Below is an attachment containing the solution.
Answer:
Explanation:
a ) Thermal efficiency = work output / heat input
= .38 MW / 1 MW = .38
OR 38%
Heat rejected at cold reservoir = heat input - work output
1 MW - .38 MW
= 0.62 MW.
b ) For reversible power output
efficiency = T₂ - T₁ / T₂ ; T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.
= 1200 - 300 / 1200 = 900 / 1200
= .75
or 75%
rate at which heat is rejected
= 1 - .75 x 1
= .25 MW .
