Answer:
The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.
Explanation:
In this ideal engine, none of these events would happen due to the nature of the efficiency.
We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.
n=W, total/(E, available).
However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.
This question is not complete.
The complete question is as follows:
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?
Explanation:
a. Using the expression;
T = 2π√R/g
where R = radius of the space = diameter/2
R = 800/2 = 400m
g= acceleration due to gravity = 9.8m/s^2
1/T = number of revolutions per second
T = 2π√R/g
T = 2 x 3.14 x √400/9.8
T = 6.28 x 6.39 = 40.13
1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute
Answer:
(4xy+5ab)(4xy-5ab)
Explanation:
16
-25
4^2 is 16 and 5^2 is 25,
Also, (x-a)(x+a) = x^2-a^2
So, this factorized is:
(4xy+5ab)(4xy-5ab)
Hope this helps!
Answer:7 cm/s
Explanation:
Given
Particle move along curve
As it reaches the (2,3) its y coordinate is increasing at 14 cm/s
Differentiating y w.r.t time
Now at (2,3)
Answer:
W = 3.12 J
Explanation:
Given the volume is 1.50*10^-3 m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:
β = 69*10^-6 (°C)^-1 V = 1.50*10^-3 m^3 ΔT = 298°C
So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):
So ΔV = 3.0843*10^-5 m^3
Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa
To get work, multiply the air pressure and the volume change.
W = 3.12 J
Hope this helps!
