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Sati [7]
4 years ago
5

Two charged bodies exert a force 0.8N on each other in a medium of relative permittivity 0.45.what will be the new force when th

ey are placed in a medium of relative permittivity 0.9,assuming all other parameters remain unchanged.​

Physics
1 answer:
Damm [24]4 years ago
7 0

The electric force between charged bodies decreases by a factor equal to the relative permittivity, ie:

F ∝ 1/ε

F = electric force, ε = relative permittivity

Find the scaling factor of F due to the change in ε:

k = 1/(0.9/0.45) = 1/2 = 0.5

Multiply the original F by k to find the new F:

F = 0.8N×k

F = 0.8N×0.5

F = 0.4N

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Convert 50cm to metre​
laila [671]

Answer:

0.5m

Explanation:

1m = 100cm 1×100 will give that answer so in converting 50cm into m you just need to do it vice versa ÷100 that will give you your answer which is 0.5m

3 0
3 years ago
calculate the acceleration of Airplane that starts at rest and reaches the speed of 45M/S in nine seconds
andreyandreev [35.5K]

Acceleration = (Speed or velocity) / Time

= 45/9

= 5 m/s²

5 0
4 years ago
Jane puts some water into an electric kettle and then she connects it to the power source. She observes that after some time the
Setler [38]
C) electrical energy is transformed into heat energy
4 0
3 years ago
Read 2 more answers
What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?
iren2701 [21]

Answer:

  • 4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

V(r) = \frac{4}{3} \ \pi \ r^3

If we got an uncertainty \Delta r the formula for the uncertainty of V is:

\Delta V(r) = \sqrt{  (\frac{dV}{dr} \Delta r)^2  }

We can calculate this uncertainty, first we obtain the derivative:

\frac{dV}{dr}  = 3 * \frac{4}{3} \ \pi \ r^2

\frac{dV}{dr}  = 4 \ \pi \ r^2

And using it in the formula:

\Delta V(r) = \sqrt{  (4 \ \pi \ r^2\Delta r)^2  }

\Delta V(r) = \sqrt{  4^2 \ \pi^2 \ r^4 \Delta r^2  }

\Delta V(r) =  4 \  \pi \ r^2 \Delta r

The relative uncertainty is:

\frac{\Delta V(r)}{V(r)}

\frac{ 4 \  \pi \ r^2 \Delta r  }{ \frac{4}{3} \ \pi \ r^3}

\frac{ 3  \Delta r  }{  r}

Using the values for the problem:

\frac{ 3 * 0.09 m  }{  5.66 m} = 0.0477

This is, a percent uncertainty of 4.77 %

4 0
3 years ago
An AC generator has 80 rectangular loops on its armature. Each loop is 12 cm long and 8 cm wide. The armature rotates at 1200 rp
Sauron [17]

Answer: 28.96 V

Explanation:

Given

No of loops on the armature, N = 80

Length of the loop, l = 12 cm = 0.12 m

Width of the loop, b = 8 cm = 0.08 m

Speed of the armature, 1200 rpm

Magnetic field of the loop, B = 0.30 T

To solve this, we use the formula

V(max) = NBAω

Where,

A = area of loop

A = l*b = 0.12 * 0.08

A = 0.0096 m²

ω = 1200 rpm = 1200 * 2π/60 rad/s

ω = (1200 * 2 * 3.142) / 60

ω = 7540.8 / 60

ω = 125.68 rad/s

Substituting the values into the formula

V(max) = NBAω

V(max) = 80 * 0.30 * 0.0096 * 125.68

V(max) = 80 * 0.362

V(max) = 28.96 V

Therefore, the maximum output voltage of the generator would be 28.96 V

5 0
3 years ago
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