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Firdavs [7]
2 years ago
9

If the magnet is now pushed into the coil and then held stationary inside the coil describe fully what would you observe on the

galvaromoter?​
Physics
1 answer:
alexira [117]2 years ago
8 0

Answer:

E = q V B     describes the electric field induced

E Proportional to V B

while the magnet is pushed into the coil the induced field (B) will increase (consider 1 turn of the coil)

If V is constant the E-field will increase due to increasing B and the galvanometer will deflect accordingly

When V drops to zero the deflection must again be zero

So one would see a blip due to the deflection of the galvanometer

Note that as V increases the galvanometer will deflect one way and then as V drops to zero the deflection will be opposite (drop to zero when V is zero)

B always increases to a constant value because of the properties of the magnet.

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Explanation:

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How was the scientific revolution important to the enlightenment
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Answer:  The scientific revolution laid the foundations for the Age of Enlightenment, which centered on reason as the primary source of authority and legitimacy, and emphasized the importance of the scientific method.

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What is variable velocity?​
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7 0
2 years ago
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To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

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