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Kobotan [32]
2 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]2 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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Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: 4.3\cdot 10^{-15} J

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

V(r)=\frac{kq}{r}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

q_1=+2.00\mu C=+2.00\cdot 10^{-6}C

and is located at the origin (x=0, y=0)

Charge 2 is

q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

r_{1A}=0.40 m

So the potential due to charge 2 is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V

The distance of point A from charge 2 is

r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m

So the potential due to charge 1 is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V

Therefore, the net potential at point A is

V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m

So the potential due to charge 1 at point B is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V

The distance of charge 2 from point B is

r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m

So the potential due to charge 2 at point B is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V

Therefore, the net potential at point B is

V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V

So the potential difference is

V_B-V_A=-29,500 V-(-2700 V)=-26,800 V

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

W=q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

In this problem, we have:

q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-26,800 V is the potential difference

Therefore, the work required on the electron is

W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

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3 years ago
A specimen of commercially pure copper has a strength of 240 MPa. Estimate its average grain diameter using the Hall-Petch equat
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Answer:

3.115× 10^{-3} meter

Explanation:

hall-petch constant for copper is given by

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now according to hall-petch equation

S_Y=S_0 +\frac{K}{\sqrt{D}}

240=25+\frac{0.12}{\sqrt{D}}

D=3.115× 10^{-3} meter

so the grain diameter using the hall-petch equation=3.115×  10^{-3} meter

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Answer:

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Answer:

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D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed

Explanation:

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2 years ago
John wants to construct a device using quartz crystal, Which device can he construct?
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Answer: Option D, piezoelectric pressure guage

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4 0
3 years ago
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