Answer:
(1) The maximum air temperature is 1383.002 K
(2) The rate of heat addition is 215.5 kW
Explanation:
T₁ = 17 + 273.15 = 290.15

T₂ = 290.15 × 3.17767 = 922.00139

Therefore,
T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K
The maximum air temperature = T₃ = 1383.002 K
(2)


Therefore;


Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg
Heat rejected per kilogram is given by the following relation;
= 0.718×(511.859 - 290.15) = 159.187 kJ/kg
The efficiency is given by the following relation;

Where:
β = Cut off ratio
Plugging in the values, we get;

Therefore;


Heat supplied = 
Therefore, heat supplied = 215491.064 W
Heat supplied ≈ 215.5 kW
The rate of heat addition = 215.5 kW.
A pendulum is probably the most common showing of this example. As the pendulum swings down, it converts its potential energy (height) into kinetic energy (velocity). At the lowest point the kinetic energy is the highest and the potential is the lowest. At the highest point in its swing the velocity is zero so the kinetic energy is zero and the potential energy is at a maximum (greatest height).
Answer:
630.75 j
Explanation:
from the question we have the following
total mass (m) = 54.5 kg
initial speed (Vi) = 1.4 m/s
final speed (Vf) = 6.6 m/s
frictional force (FF) = 41 N
height of slope (h) = 2.1 m
length of slope (d) = 12.4 m
acceleration due to gravity (g) = 9.8 m/s^2
work done (wd) = ?
- we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy
wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)
wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)
where wd = work done
m = mass
h = height
g = acceleration due to gravity
FF = frictional force
d = distance
Vf and Vi = final and initial velocity
wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)
wd = 630.75 j
Answer:
2.605m
Explanation:
Using the formula for calculating Range (distance travelled in horizontal direction)
Range R = U√2H/g
U is the speed = 4.8m/s
H is the maximum height = ?
g is the acc due to gravity = 9.8m/s²
R = 3.5m
Substitute into the formula and get H
3.5 = 4.8√2H/9.8
3.5/4.8 = √2H/9.8
0.7292 = √2H/9.8
square both sides
0.7292² = 2H/9.8
2H = 0.7292² * 9.8
2H = 5.21
H = 5.21/2
H = 2.605m
Hence the height of the ball from the ground is 2.605m
I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!