Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
Answer:height above ground at which projectile have velocity
0.5v is (0.0375v^2)
Explanation:
Using Vf = Vi - gt
Where Vf is final velocity
Vi is initial velocity
g is the acceleration due to gravity
t is the time taken
So, 0.5v = v - gt
t = 0.05v
Therefore height h = vt - 0.5gt^2
Subtitute t
h = 0.05v^2 - 0.0125v^2
h = 0.0375v^2
Hi , the answer is D , battery.
"Energy and Momentum" is always conserved in an inelastic condition
Hope this helps!
