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nadya68 [22]
3 years ago
14

If an object is placed at distance of 16cm from a plane mirror, How far would it be from it's image?​

Physics
1 answer:
Lelechka [254]3 years ago
3 0

Explanation:

A plane mirror always creates an image with the same distance to the mirror as the object, only in the other direction. So both of them have a distance of 10cm, one is 10cm to the left, one 10cm to the right, thus their mutual distance is 20cm

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I think hes cool

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Can you guys please help me with this science quiz
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1 is 90, 2 is 200 and 3 is 5

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im big brain so i know lol

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there is a fish called an archer fish that shoots drops of water at insects resting on branches above the water. If the Archer f
Mariulka [41]

Answer:

=3.5 m/s

Explanation:

y = x tanθ - 1/2 g x² / (u²cos²θ )

y = 0.25 , x = 0.5, θ = 40°

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4 0
3 years ago
n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet
enot [183]

Answer:

The  corresponding  magnetic field is  

Explanation:

From the question we are told that

    The electric field amplitude is  E_o   =  611\  V/m

   

Generally the  magnetic  field amplitude is  mathematically represented as

              B_o  =  \frac{E_o }{c }

Where c is the speed of light with a constant value

         c = 3.0 *0^{8} \ m/s

So  

        B_o   =  \frac{611 }{3.0*10^{8}}

         B_o   =  2.0 4 *10^{-6} \  Vm^{-2} s

Since 1  T  is  equivalent to  V  m^{-2} \cdot  s

         B_o  =  2.0 4 *10^{-6} \ T

6 0
4 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
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