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In order to compute the torque required, we may apply Newton's second law for circular motion:
Torque = moment of inertia * angular acceleration
For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds
α = (3.5 - 0) / 1.8
α = 1.94 rad/s²
The moment of inertia of a thin disc is given by:
I = MR²/2
I = (0.21*0.1525²)/2
I = 0.002
τ = 1.94 * 0.002
τ = 0.004
The torque is 0.004
Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
Answer:
(a) 62.5 m
(b) 7.14 s
Explanation:
initial speed, u = 35 m/s
g = 9.8 m/s^2
(a) Let the rocket raises upto height h and at maximum height the speed is zero.
Use third equation of motion
h = 62.5 m
Thus, the rocket goes upto a height of 62.5 m.
(b) Let the rocket takes time t to reach to maximum height.
By use of first equation of motion
v = u + at
0 = 35 - 9.8 t
t = 3.57 s
The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.
