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3 years ago

A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase

Physics

1 answer:

SashulF [63]3 years ago

5 0

Answer:

E_Phase = 560V

Explanation:

The computation of the voltage i.e. dropped across each phase is shown below:

Given that

The delta connection line voltage is

E_line = 560 V

And, in the case of delta connection, the line voltage would be equivalent to the phase voltage

That means

E_Phase = E_Line

= 560 V

Hence, the voltage i.e. dropped across each phase is

E_Phase = 560V

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3 years ago

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the

Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

Given:

Initial angular velocity = ωi = 2.70 rad/s

Final angular velocity = ωf = -2.70 rad/s (negative sign is

due to the movement in opposite direction)

Change in time period = Δt = 1.50 s

Required:

Change in angular velocity = Δω = ?

Average angular acceleration = αav = ?

Solution:

Angular velocity (Δω):

Δω = ωf - ωi

Δω = -2.70 - 2.70

Δω = -5.4 rad/s.

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αav = Δω/Δt

αav = -5.4/1.50

αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

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3 years ago

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How do scientists classify species?

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Answer:

D. By comparing traits

Explanation:

Because age isn't genetic, as well as names, as well as who discovered, but traits are genetic.

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3 years ago

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At the equator earth rotates with a velocity of about 465 m/s.

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The given velocity is 465 m/s.

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Part b.
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3 years ago

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A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

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3 years ago