From a to b speed is 600+40 = 640
from b to a speed is 600-40 = 560
let t be the number of hours of flight. This would mean it would have traveled a distance of 640 miles and the distance yet to travel is 2400-640t
Time left will be (2400-640t)/640. But if they were to return to a it would fly 640t miles at 560mph which will take (640t/560) hrs
(2400-640t) / 640 = 640t / 560
560(2400 - 640t) = 640t x 640
t = 1.75hrs
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed,
rev/sec
rad/sec
= 5.89 rad/sec
Therefore, force required,

= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
Answer:
no picture or anything so cant anwser
Explanation:
consider the motion in x-direction
= initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m
= acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation
X =
t + (0.5)
t²
100 =
(4.60)
= 21.7 m/s
consider the motion along y-direction
= initial velocity in y-direction = ?
Y = vertical displacement = 0 m
= acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation
Y =
t + (0.5)
t²
0 =
(4.60) + (0.5) (- 9.8) (4.60)²
= 22.54 m/s
initial velocity is given as
= sqrt((
)² + (
)²)
= sqrt((21.7)² + (22.54)²) = 31.3 m/s
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg