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3 years ago

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Need help pls and explain also

1 answer:

Alisiya [41]3 years ago

3 0

This should be the answer

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The inductive reactance of the circuit is exactly twice the resistance: XL=2R. Adjust the phasor that represents the voltage acr

SVETLANKA909090 [29]

Answer:

∅= $63.43^{0}$

Explanation:

z=impedance

$x_{l}$ =2R

R=R

The resultant of the resistances in the circuit is called impedance

$x_{l}$ is inductive reactance of the circuit

R is the resistance of the resistor

z= $\sqrt{xl^{2}+R^2 }$

z= $\sqrt{2R^{2}+R^2 }$

Z= $\sqrt{5R^2}$

Z=R $\sqrt{5}$ ohms

tan∅=2R/R

tan∅=2

∅=Tan^-1(2)

∅= $63.43^{0}$

phase angle is ∅= $63.43^{0}$

3 0

3 years ago

What is non uniform motion?

Dmitriy789 [7]

Explanation:

If a body does not cover a equal distance at a equal interval of time it is said to be non uniform motion...

$...$

4 0

2 years ago

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o

zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

$F= ma$

$a=\dfrac{F}{m}$

Put the value of F

$a=\dfrac{kq^2}{mr^2}$

Put the value into the formula

$414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}$

$q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}$

$q^2=1.84\times10^{-14}$

$q=0.135\times10^{-6}\ C$

$q=0.135\ \mu C$

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0

3 years ago

Read 2 more answers

I need a sentence for the word sandy soil

rusak2 [61]

One day, as I was walking, I found some sandy soil beside the road.

6 0

3 years ago

Read 2 more answers

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.

wlad13 [49]

Answer:

12m/s

Explanation:

$v_f=v_o+at$

Let's call the velocity that the car maintains for 10 seconds $v_f_1$ , and the final velocity $v_f_2$ .

$v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s$

Hope this helps!

5 0

3 years ago