The force exerted on the car during this stop is 6975N
<u>Explanation:</u>
Given-
Mass, m = 930kg
Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s
Time, t = 2s
Force, F = ?
F = m X a
F = m X s/t
F = 930 X 15/2
F = 6975N
Therefore, the force exerted on the car during this stop is 6975N
Answer:
2046.37 kPa
Explanation:
Given:
Number of moles, n = 125
Temperature, T = 20° C = 20 + 273 = 293 K
Radius of the cylinder, r = 17 cm = 0.17 m
Height of the cylinder, h = 1.64 m
thus,
volume of the cylinder, V = πr²h
= π × 0.17² × 1.64
= 0.148 m³
Now,
From the ideal gas law
we have
PV = nRT
here,
P is the pressure
R is the ideal gas constant = 8.314 J / mol. K
thus,
P × 0.148 = 125 × 8.314 × 293
or
P × 0.148 = 304500.25
or
P = 2046372.64 Pa = 2046.37 kPa
Well this question looks like it makes some assumptions. So assuming that both cars have the same mass and experience the same wind resistance regardless of speed and same internal frictions, then we could say "The car that finishes last has the lowest power". The reason is that for a given race the cars must overcome losses associated with motion. Since they all travel the same distance, the amount of work will be the same for both. This is because work is force times distance. If the force applied is the same in both cases (identical cars with constant wind resistance) and the distance is the same for both (a fair race track) then W=F·d will be the same.
Power, however, is the work done divided by the time over which it is done. So for a slower car, time t will be larger. The power ratio W/t will be smaller for the longer time (slower car).
The 1st one goes two added sodoes the second one then the third goes to removed then the fourth goes to added and the rest go to removed