Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.
Answer:
Frictional force increases with the increase in the roughness of the surface.
Explanation:
You will see that the rougher the surface, the greater the wear and tear.
part 1
mass = ρ x V
mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg
PE (potential energy)= mgh
PE = 6608.2 kg x 9.81 x 403
PE = 2.61 x 10⁷ J
part 2
megaton of TNT (Mt) =4.2 x 10¹⁵ J
convert PE to Mt:
2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt
Answer:
The object starts away from the origin and then moves toward the origin at a constant velocity. Next, it stops for one second. Finally, it moves away from the origin at a greater constant velocity.