Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
Answer:
24.3 m/s
Explanation:
1 kmh = 0.27 m/s, that makes a conversion ratio of 0.27/1kmh
x 
The "kmh" n the top and bottom cancel out. And then you just multiply the top 90 x 0.27 and the bottom 1 x 1 to get
and since its over 1 its just 24.3 m/s
Answer:
(a) v = 3..6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
Explanation:
from the question we have the following:
mass of the car (Mc) = 24,000 kg
initial velocity of the car (u) = 4 m/s
mass of water (Mw) = 3000 kg
final velocity of the car (v) = ?
(a) we can calculate the final momentum of the car by applying the conservation of momentum where
initial momentum = final momentum
Mc x U = (Mc + Mw) x V
24000 x 4 = (24000 + 3000) x v
96,000 = 27000v
v =3.6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
