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Kobotan [32]
3 years ago
5

A particular motor rotates at 3000 revolutions per minute. What is its speed in rad/sec, and how many seconds does it takes to m

ake one revolution?
Engineering
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

\omega=\dfrac{2\pi N}{60}\ rad/s

Now by putting the values in the above equation

\omega=\dfrac{2\pi \times 3000}{60}\ rad/s

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

\omega=\dfrac{ 3000}{60}\ rad/s

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

\dfrac{1}{50}=0.02\ s

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Solution :

Given :

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Thickness, t = 20 mm

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                                 = 460 mm

So, internal radius, r = 230 mm

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Density of molten metal, ρ = $7.2 \ g/cm^3$

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The height of pouring cavity above parting surface is h = 300 mm

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So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

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            $=0.3324 \ m^2$

$F=\rho g A h$

   $=7200 \times 9.81 \times 0.3324 \times 0.3$

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3 years ago
a coil consists of 200 turns of copper wire and has a cross-sectional area of 0.8mm square . The mean length per turn is 80 cm a
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3 years ago
A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given
viktelen [127]

Answer:

A.) P = 2bar, W = - 12kJ

B.) P = 0.8 bar, W = - 7.3 kJ

C.) P = 0.608 bar, W = - 6.4kJ

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That is, P1V1^n = P2V2^n

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the final pressure P2 = 2 bar. 

A.) When n = 0

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(V2/V1)^0 = 1

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Work = P1V1 × ln ( V2/V1 )

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Work = (2×10^5×0.04) - (0.608 10^5×0.1)/ ( 1 - 1.3 )

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5 0
4 years ago
When the compression process is non-quasi-equilibrium, the molecules before the piston face cannot escape fast enough, forming a
muminat

Answer:

a. true

Explanation:

Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times.  When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.

While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.

5 0
3 years ago
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