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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 9.13e+3 kg railroad car is rolling at 3.15 m/s when a 4.20e+3 kg load of gravel is suddenly dropped in from directly above. Wh
Feliz [49]

Answer:

6.85 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, since there are no external forces acting, the total momentum before and after must be conserved. So we can write:

m_1 v_1 = m_2 v_2

where

m_1 = 9.13\cdot 10^3 kg is the initial mass of the car

v_1 = 3.15 m/s is the initial speed of the car

m_2 = 9.13\cdot 10^3 kg - 4.20\cdot 10^3 kg=4.93\cdot 10^3 kg is the mass of the car after the load of gravel is dropped

v2 is the final speed of the car

Solving for v2, we find

v_2 = \frac{m_1 v_1}{m_2}=\frac{(9.13\cdot 10^3)(3.15 m/s)}{4.93\cdot 10^3}=6.85 m/s

6 0
3 years ago
During most of its lifetime, s star maintains an equilibrium size in which the inward force of gravity on each atom is balanced
frez [133]

Answer:

r_f=137493m

v=8638940m/s

Explanation:

During this process the mass M=2\times10^{30}Kg will be considered constant. We start from a radius r_i=7\times10^8m and a period T_i=30\ days=(30)(24)(60)(60)s=2592000s. The final period is T_f=0.1s.

Angular momentum <em>L</em> is conserved in this process. We can use the formula L=I\omega, where I is the momentum of inertia (which for a solid sphere is I=\frac{2mr^2}{5}) and \omega=\frac{2\pi }{T} is the angular velocity, so we can write the star's angular momentum as:

L=I\omega=\frac{2mr^2}{5}\frac{2\pi }{T}=\frac{4\pi mr^2 }{5T}

Since L_f=L_i we have:

\frac{4\pi mr_f^2 }{5T_f}=\frac{4\pi mr_i^2 }{5T_i}

Which can be simplified as:

\frac{r_f^2 }{T_f}=\frac{r_i^2 }{T_i}

Which means:

r_f=\sqrt{\frac{r_i^2 T_f}{T_i}}=r_i \sqrt{\frac{T_f}{T_i}}

Which for our values is:

r_f=r_i \sqrt{\frac{T_f}{T_i}}=(7\times10^8m) \sqrt{\frac{0.1s}{2592000s}}=137493m

And we calculate the speed of a point on the equator by dividing the final circumference over the final period:

v=\frac{C_f}{T_f}=\frac{2\pi r_f}{T_f}=\frac{2\pi (137493m)}{(0.1s)}=8638940m/s

3 0
4 years ago
In August 2011, the Juno spacecraft was launched from Earth with the mission of orbiting Jupiter in 2016. The closest distance b
Vikentia [17]

(a) 8927 mi/h

In order to calculate the average speed, we need to convert the time (t=5.0 y) into hours first. In 1 year, we have 365 days, each day consisting of 24 hours, so the time taken is:

t=(5.0 y)(365 d/y)(24 h/d)=43,800 h

The distance covered by the spacecraft is

d=391 mil. mi = 391\cdot 10^6 mi

Therefore, the average speed is just the ratio between the distance covered and the time taken:

v=\frac{d}{t}=\frac{391\cdot 10^6 mi}{43,800 h}=8,927 mi/h

(b) 35 minutes (2097 seconds)

The transmitted signals (which is a radio wave, which is an electromagnetic wave) travels back to the Earth at the speed of light:

c=3.0\cdot 10^8 m/s

Since 1 miles = 1609 metres, the distance covered  by the signal is

d=391\cdot 10^6 mi \cdot (1609 m/mi)=6.29\cdot 10^{11} m

So, the time taken by the signal will be

t=\frac{d}{v}=\frac{6.29\cdot 10^{11} m}{3.0\cdot 10^8 m/s}=2097 s

And since 1 minute = 60 sec, the time taken is

t=2097 s \cdot \frac{1}{60 s/min}\sim 35 min

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Elaborate . on what can be learned about the properties by the location of element on the periodic table . Use atomic number 13
soldier1979 [14.2K]

Aluminum is a type of metal and that is a properties by the location of element

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3 years ago
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makkiz [27]

Answer:

Yes.

Explanation:

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3 0
3 years ago
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