Answer:
a) tD = 4.36 sec
b) tD = 5.2 sec
Explanation:
a)
V = 0
work-done = change in kinetic energy
f.m.g(d) = 1/2m(55^2 - v^2)
divided both sides by m
1/2(55^2 - v^2) = f.g(d)
1/2(55 * 1.467)^2 ft^2/sec^2 = 0.6 * (9.81 * 3.28) * d
3255.03 = 19.306d
d = 3255.03 / 19.306
d = 168.6 ft
now D = 520 - 168.6 = 351.4
therefore reaction/perception time = tD
tD = 351.4 / ( 55 * 1.467)
tD = 4.36 sec
b)
Also here, V = 35 mph
so
1/2(55^2 - 35^2)*(1.467)^2 ft^2/sec^2) = 0.6 * (9.81 * 3.28) * d
1936.88 = 19.306d
d = 1936.88 / 19.306
d = 100.325ft
also D = 520ft - 100.325ft = 419.675
so tD = 419.675 / ( 55 * 1.467)
tD = 5.2 sec
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An ideal gas initially occupying 0.020 m3 at 1.0 MPa is quasistatically expanded inside a piston-cylinder device at a constant pressure until its volume doubles. Next the expansion is continued at constant volume till the pressure reaches half of the initial pressure. Finally it is brought back to the initial state in a polytropic process with exponent n=1.6
a. Draw the processes on a P-v diagram and calculate the total work.
b. Calculate the total heat transfered, what is the difference between the initial and final temperature?an answer is to present a question of how you are not able to join the world and how you can help please answer
