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svp [43]
3 years ago
7

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

. Consider the merry-go-round to be a uniform disk of radius 5.5m and mass 29,000 kg. Suppose that it is supported by bearings that produce negligible friction torque.What torque specifications must the new motor satisfy?
Physics
1 answer:
pashok25 [27]3 years ago
5 0

Answer:

109656.25 Nm

Explanation:

\omega_f = Final angular velocity = 1.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm

The torque specifications must be 109656.25 Nm

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A uniform steel plate has an area of 0.819 m2. When subjected to a temperature difference between its sides, a heat current* of
Tju [1.3M]

Answer:

ΔT / Δx = 771 K/m

ΔT = 771 x 0.0475 = 36.62 k

Explanation:

P = 31700 W, A = 0.819 m^2, Δx = 0.0475 m, K = 50.2 W /m k

Use the formula of conduction of heat

H / t = K A x ΔT / Δx

So, ΔT / Δx = P / K A

ΔT / Δx = 31700 / (50.2 x 0.819)

ΔT / Δx = 771 K/m

Now

ΔT = 771 x 0.0475 = 36.62 k

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3 years ago
how much time would it take for the sound of thunder to travel 1,500 meters if sound travels at a speed of 330m/s
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Answer:

4.545454 seconds

Explanation:

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5 0
4 years ago
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nignag [31]

We will have that the graph that describes the scenario is given by graph B.

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If you move a substance from one container to another and its volume changes, what is the substance
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4 years ago
An isolated capacitor with capacitance C = 1 µF has a charge Q = 45 µC on its plates.a) What is the energy stored in the capacit
Roman55 [17]

Answer:

a) Energy stored in the capacitor, E = 1.0125 *10^{-3} J

b) Q = 45 µC

c) C' = 1.5 μF

d)  E = 6.75 *10^{-4} J

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J

4 0
3 years ago
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