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jolli1 [7]
3 years ago
11

A spinning flywheel has rotational inertia I = 474.0 kg·m2. Its angular velocity decreases from 26.2 rad/s to zero in 204.0 s du

e to friction. What is the frictional torque acting?
Physics
1 answer:
Flura [38]3 years ago
7 0

Answer:

\tau = 60.88 N.m

Explanation:

given,

rotational inertia, I = 474.0 kg·m²

decrease in angular velocity

ω₁ = 26.2 rad/s        ω₂ = 0 rad/s

time = 204 s

torque = ?

\tau = I \alpha

\alpha = \dfrac{\omega_1-\omega_2}{t}

\alpha = \dfrac{26.2-0}{204}

   α = 0.128 rad/s²

\tau = I \alpha

\tau = 474\times 0.128

\tau = 60.88 N.m

frictional torque acting by the flywheel is equal to 60.88 N.m

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vladimir1956 [14]

Homeostasis is the state of maintaining steady internal conditions by the living organisms. Cells obtain energy through the process of cellular respiration. It is a metabolic process of conversion of the biochemical energy from the nutrients into adenosine triphosphate (ATP). It also maintains the temperature of the body. The energy produced during this process is used in the cell division and repair of the cells by the breakdown of ATP and maintains homeostasis. They exchange substances with the new cells and also eliminate the wastes thereby maintaining homeostasis.

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7 0
3 years ago
At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl
bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0
3 years ago
Forces of attraction limit the motion of particles most in
Alexandra [31]

<em><u>Answer:</u></em>

Solids

<em><u>Explanation:</u></em>

<u>Solids</u> have definite shapes and definite volumes. The forces of attraction between the molecules of a solid substance are strong and the intermolecular spaces are very small. Due to this, the motion of molecules within a solid substance are very difficult that they only vibrate in their positions.

<u>For liquids</u>, they have definite volumes and indefinite shapes. The forces of attraction between the molecules of a liquid are intermediate and the the intermolecular spaces are intermediate as well. Due to this, the motion of molecules within a liquid substance is not as difficult as it is within solids.

Finally, <u>for gases</u>, they have indefinite shapes and indefinite volumes. The forces of attraction between the molecules of a gas are weak and the intermolecular spaces are large. Due to this, the motion of particles of gas is very easy.

Hope this helps :)

7 0
3 years ago
Read 2 more answers
True or false: objects fall toward earth at a rate of 9.8 m/s because of centripetal force.
Citrus2011 [14]
Sadly, no. The statement kind of has some appropriate words in it, but it's badly corrupted. Objects don't fall to Earth at a rate of 9.8 m/s, and the force that accelerates them downward is not a centripetal one.
4 0
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If the star Alpha Centauri were moved to a distance 10 times farther than it is now, its parallax angle would
7nadin3 [17]

Answer:

decrease by a factor 10

Explanation:

The parallax angle of a close star is given by

p=\frac{1}{d}

where

p is the parallax angle

d is the distance of the star from Earth, in parsecs

From the formula we see that the parallax angle is inversely proportional to the distance.

In this problem, the distance of the star is increased by a factor 10:

d' = 10 d

so the new parallax angle would be

p'=\frac{1}{10 d}=\frac{1}{10}\frac{1}{d}=\frac{p}{10}

So, the parallax angle would decrease by a factor 10.

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3 years ago
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