Answer:
9500 kJ; 9000 Btu
Explanation:
Data:
m = 100 lb
T₁ = 25 °C
T₂ = 75 °C
Calculations:
1. Energy in kilojoules
ΔT = 75 °C - 25 °C = 50 °C = 50 K
2. Energy in British thermal units
2.18 The net potential energy between two adjacent ions, EN, may be represented by the following equation: (1) Calculate the bon
1 answer:
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<u>Explanation:</u>
5 Horsepower for 30 mins,
(5)(745.7) = 3.7285 KW power delivered
General Efficiency of IC engine = 20%
Power required =
Energy required per week,
=P × Time = 18.64 × 60 × 30 = 33.5565 MJ
Lawn area = (30) (20) = 600
let sunlight hours be 8 hours
Hence, solar power input on lawn,
=5.62×3600 = 20232 kJ//day
energy input in lawn = (600) (20232) (7)
= 84974.4 mJ/week
Chemical efficiency by photosynthesis = 4%
Chemical content in grass = (84974.4) (0.04)
= 3398.97 mJ
Mass of the clippers
pounds
Removing water content,
dried grass clippings
= 11533.25 gallons
Trash cans repaired
By burning the gas, total energy input = 3398.97 MJ × 0.2
= 679.794 MJ
Efficiency of steeling engine = 20%
Energy output by engine = 679.794 ×0.2
= 135.96 mJ
Energy required by mover = 33.5565 mJ
Hence, Energy (output) ⇒ energy required
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.