<span>When the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.
Given:
a = 86 m/s^2
t = 1.7 s
Solution:
d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.7) + 0.5 (86) (1.7)^2
d = 124.27 m
vf = vi + at
vf = 0 + 86 (1.7)
vf = 146.2 m/s (velocity when the fuel is consumed completely)
Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 146.2 + (-9.8) (t)
t = 14.92 s
Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2)
d = 1090.53 m
Then, we determine the maximum altitude:
d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m</span>
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.
Answer:
Average speed of car in the first trip is 10 km/hr
Explanation:
It is given that first the car drives 6 hours to the east
Then travels 12 km to west in 3 hours
Average speed for the entire trip = 8 km/hr
Total time = 3+6 = 9 hour
So distance traveled in 9 hour = 9×8 = 72 km
As the car travel 12 km in west so distance traveled in east = 72-12 = 60 km
Time by which car traveled in east = 6 hour
So speed
So average speed of car in the first trip is 10 km/hr