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2 years ago

Need help please in physics

Physics

1 answer:

AnnZ [28]2 years ago

3 0

Answer:

the switching circuitry is opened the soft iron armature is replaced to tasty le macha the switching circuitry is closed explain what happens when the switch circuit is a wonder you know about the characteristics of open to close this which right take the example of an electric generator for an electric motor which rotates in half rotation and change the armature and just try it ok I am sorry

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If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, how

taurus [48]

T=2s
g=10m/s2
h=?
free fall: h=gt2/2
              h= 10*4/2
               h=40/2
              h=20m

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2 years ago

1. What is the total distance traveled?<br> A 3.0m <br> B 4.0m <br> C 5.0m <br> D 6.0m

Ostrovityanka [42]

Answer:

Explanation:

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2 years ago

Which component of a galaxy is often found between the stars and looks like a cloud or smoke?

slava [35]

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1 year ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

2 years ago

an astronaut weighs 104 newtons on the moon where the strength of gravity is 1.6 newtons per kilogram what is her mass

Andreas93 [3]

M = W/g
mass (m)
weight (W) and strength of gravity (g)
Therefore the mass of the astronaut is 65 kilograms

7 0

2 years ago

Need help please in physics ​

Need help please in physics