Answer:
(b) 2.40 x kg/s
Explanation:
Given that: Total mass of the rocket = 3.003 x kg
acceleration of the rocket = 36.0 m
speed of the exhausted gases = 4.503 x m/s
Rate at which rocket was initially burning fuel =
But,
time =
=
= 125.0833 s
So that;
Rate at which rocket was initially burning fuel =
= 2400.8001
= 2.40 x kg/s
Therefore, the initial rate at which the rocket burn fuel is 2.40 x kg/s.
And walk another 25m west
Answer:
Answer:
77.58°
Explanation:
Let the projectile is projected at an angle θ and the velocity of projection is u.
At the maximum height, the projectile has only horizontal component of velocity.
Let the velocity at maximum height is v.
According to the question, the velocity at maximum height is 21.5% of initial velocity.
v = 21.5 % of u
u Cos θ = 21.5 % of u
u Cos θ = 0.215 u
Cos θ = 0.215
θ = 77.58°
Thus, the angle of projection is 77.58°.
Explanation:
Answer:
ωf = 4.53 rad/s
Explanation:
By conservation of the angular momentum:
Ib*ωb = (Ib + Ic)*ωf
Where
Ib is the inertia of the ball
ωb is the initial angular velocity of the ball
Ic is the inertia of the catcher
ωf is the final angular velocity of the system
We need to calculate first Ib, Ic, ωb:
ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s
Now, ωf will be: