Height of the waterfall is 0.449 m
its horizontal distance will be 2.1 m
now let say his speed is v with which he jumped out so here the two components of his velocity will be


here the acceleration due to gravity is 9.81 m/s^2 downwards
now we can find the time to reach the other end by y direction displacement equation


also from x direction we can say


now we have

we will plug in this value into first equation



now as we know that

t = 0.63 s


so his minimum speed of jump is 4.1 m/s
Answer:
It's (c). ..........................
Answer:
1.3823 rad/s
20.7345 m/s
28.66129935 m/s²
2006.29095 N radially outward
Explanation:
r = Radius = 15 m
m = Mass of person = 70 kg
g = Acceleration due to gravity = 9.81 m/s²
Angular velocity is given by
Angular velocity is 1.3823 rad/s
Linear velocity is given by
The linear velocity is 20.7345 m/s
Centripetal acceleration is given by
The centripetal acceleration is 28.66129935 m/s²
Acceleration in terms of g
Centripetal force is given by
The centripetal force is 2006.29095 N radially outward
The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.
Answer:
18.43m
Explanation:
At the initial stage, the car accelerates at 2.0 m/s² for 6.7 seconds, to get the velocity of the car at this point, we will use the equation of motion;
v = u + at
v = 0+(2.0)(6.7)
v = 13.4m/s
The velocity of the car during this time is 13.4m/s.
Also, if the car slows down at a rate of 1.5m/s², we can also calculate the time it took for the car to decelerate (check the attachment for diagram).
v = u+at
v = 0
u = 13.4m/s
a = -1.5m/s² (deceleration is negative acceleration)
0 = 13.4+-1,5t
-13.4 = -1.5t
t = 13.4/1.5
t = 8.93s
Hence it took the car 8.93s to slow down for the next stop sign.
To calculate how far apart the stop signs are, we need to calculate the total distance AD according to the diagram
Distance covered = AD = 18.43m
HENCE THE STOP SIGNS ARE 18.43m apart