The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is
∑ F = F[a] - F[f] - F[air] = ma
3100 N - 200 N - F[air] = (650 kg) (3 m/s²)
Solve for F[air] :
F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)
F[air] = 3100 N - 200 N - 1950 N
F[air] = 950 N
<span>Pice=920kg/m^3
deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa
P=Po + deltaP=101.325 + 9022 = 9123kPa</span>
Explanation:
There are three forces on the bicycle:
Reaction force Rp pushing up at P,
Reaction force Rq pushing up at Q,
Weight force mg pulling down at O.
There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.
Sum of the forces in the y direction:
Rp + Rq − (15)(9.8) = 0
Rp + Rq − 147 = 0
Sum of the moments at P:
(15)(9.8)(0.30) − Rq(1) = 0
44.1 − Rq = 0
Sum of the moments at Q:
Rp(1) − (15)(9.8)(0.70) = 0
Rp − 102.9 = 0
Sum of the moments at O:
Rp(0.30) − Rq(0.70) = 0
0.3 Rp − 0.7 Rq = 0
Any combination of these equations will work.
Answer: It eventually loses its stamina, and would come to a stop.
Explanation:
