Answer: 34.4 sec
Explanation: Assuming that the desacceleration was constant, we have a = -1.5m/s²
vi = 186 km/h → 51.6 m/s
vf = 0 (because it stopped)
Vf = Vi + at
0 = 51.6 - 1.5t
1.5t = 51.6
t = 34.4 s
<span>P1V1/T1=P2V2/T2
isolate V2 and calculate
remember the temperature has to be in Kelvin
K=C+273
the pressure in this case is not necessary to convert to atm you can use the mmHg you will get the same answer</span>
Given that,
Radius of track, r = 50 m
time , t = 9 s
velocity, v = ?
Distance covered by car in one lap around a track is equal to the circumference of the track.
C = 2 π r = 2 * 3.14 * 50
C = 314.159 m
Distance covered by car, s = 314.159 m
Velocity = distance/ time
V = 314.159 / 9
V = 34.9 m/s
The average velocity of car is 34.9 m/s.
Answer:
C: Net Torque divided by Moment of Inertia
Explanation:
Torque = Moment of Inertia x Angular Acceleration.
The final velocity, the initial velocity, and the time :)