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NISA [10]
3 years ago
8

It has been suggested that rotating cylinders about 10 mi long and 5.0 mi in diameter be placed in space and used as colonies. W

hat angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth
Physics
1 answer:
Digiron [165]3 years ago
7 0

Centripetal acceleration is given by:

a = v²/r

a = centripetal acceleration, v = speed of cylinder at surface, r = radius of cylinder

This equation relates the speed of the cylinder at its surface to its angular velocity:

v = rω

v = speed at surface, r = radius, ω = angular velocity

Make a substitution:

a = (rω)²/r

a = rω²

Given values:

a = 9.81m/s², r = (5.0mi)/2 = 2.5mi = 4023m

Plug in and solve for ω:

9.81 = 4023ω²

ω = 0.049rad/s

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A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit
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Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

So, the magnitude of angular acceleration is 232.38\ rad/s^2. Hence, this is the required solution.

6 0
3 years ago
A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu
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Answer:

The work done is 5136.88 J.

Explanation:

Given that,

n = 1.90 mol

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If the initial volume is V then the final volume will be V/3.

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Using formula of work done

W=nRT\ ln(\dfrac{V_{f}}{V_{i}})

Put the value into the formula

W=1.90\times8.314\times296\ ln(\dfrac{\dfrac{V}{3}}{V})

W=1.90\times8.314\times296\ ln(\dfrac{1}{3})

W=−5136.88\ J

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Hence, The work done is 5136.88 J.

5 0
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_{94}^{239}\textrm{Pu}+_2^4\textrm{He}\rightarrow _{96}^{242}\textrm{Cm}+_0^\textrm{n}

Nuclear fission is a process which involves the conversion of a heavier nuclei into two or more small and stable nuclei along with the release of energy.

_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{36}^{91}\textrm{Kr}+_{56}^{142}\textrm{Ba}+_{0}^{1}\textrm{n}

8 0
4 years ago
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