<u>Given that</u>
mass (m) = 1300 Kg ,
height (h) = 1500 m
Determine the potential energy ?
P.E = m × g × h
= 1300 × 9.81 × 1500
= 19129500 Joules
= 19129.5 KJ
Answer: -2 km
Explanation:
If we imagine Jin's movement to be the hypothenuse of a right triangle, then the southern component of Jin's movement corresponds to the side of the triangle opposite to the angle of 30 degrees. Therefore, the magnitude of this southern component is given by
However, the angle of 30 degrees is south of east: this means that the direction of this southern component is south, and since we generally take north as positive direction, we must add a negative sign, so the correct answer is
-2 km
Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>
<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>
<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>
<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
Answer:b)1770 kWh
Explanation:
Given
volume of water 
Temperature rise 
also 1 kg mass is approximately is 1 gallon
therefore 40,000 gallon is equivalent to 3.8\times 40000 kg
heat Required to raise temperature is
Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
