Answer: 2933 kg
Explanation:
Given
From the law of conservation of linear momentum, we know that
m(i)v(i) = m(f)v(f), where there is no external force acting on the system
m(i) = initial mass of the freight car system
m(f) = maximum mass of the freight car system
v(i) = initial linear velocity of the system
v(f) = final linear velocity of the system
2000 * 4.4 = m(f) * 3
3m(f) = 8800
m(f) = 8800 / 3
m(f) = 2933.3 kg
Therefore the maximum mass of grain that it can accept is 2933 kg
Answer:
KE (5) = 1/2 M V^2 = 25/2 M
KE (10) = 1/2 M V^2 = 100/2 M
KE (10) - KE (5) = M/2 (100 - 25) = 75/2 / M
An object traveling at 10 m/s has 4 times the kinetic energy as an object traveling at 5 m/s
Total work would depend on the mass being accelerated
d = 400 m (8 laps) = 2400 m
t = 14.5 min
v = 0
average t = 14.5 min (60 sec per min) = 870 s/8 laps = 108.75 s per lap
speed = distance /time
400/108.75 sec = 3.67 m/s
I=22 ÷20
Explanation:
then put the on the calculator you'll get you answer
Answer:
1 second later the vehicle's velocity will be:
5 seconds later the vehicle's velocity will be:
Explanation:
Recall the formula for the velocity of an object under constant accelerated motion (with acceleration ""):
Therefore, in this case and
so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression: