Answer:
W= 4.89 KJ
Explanation:
Lets take
temperature of hot water T₁ = 100⁰C
T₁ = 373 K
Temperature of cold ice T₂= 0⁰C
T₂ = 273 K
The latent heat of ice LH= 334 KJ
The heat rejected by the engine Q= m .LH
Q₂= 0.04 x 334
Q₂= 13.36 KJ
Heat gain by engine = Q₁
For Carnot engine
Q₁ = 18.25 KJ
The work W= Q₁ - Q₂
W= 18.25 - 13.36 KJ
W= 4.89 KJ
Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
dh/dt = 64/5π(1.5)³
dh/dt = 1.21 in/sec
