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
A. Through the direct contact of particles
<u>Answer</u>
8. 2 Hz
9. 0.5 seconds
10. 20 cm
<u>Explanation</u>
<u>Q 8</u>
Frequency is the number of oscillation in a unit time. It is the rate at which something repeats itself in a second.
In this case, the spring bob up and down 2 times per second.
∴ Frequency = 2 Hz
<u>Q 9</u>
Period is the time taken to complete one oscillation.
2 oscillations takes 1 second
1 oscillation = 1/2 seconds.
∴ Period = 0.5 seconds
<u>Q 10</u>
Amplitude is the the maximum displacement of the spring.
In this case the spring bob up 20 cm. This is it's displacement.
∴ Amplitude = 20 cm
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
