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Masja [62]
3 years ago
11

The electric field, generated by a point charge, has strength of 180,000 N/C at a point 1.50 cm from the point charge. What is t

he charge, if the field points towards the charge
Physics
1 answer:
cestrela7 [59]3 years ago
7 0

Given Information:  

Electric field = E =  180,000 N/C

Distance = r = 1.50 cm = 0.015 m

Required Information:  

Charge = q = ?

Answer:

charge = q = 4.5x10⁻⁹ C

Explanation:

The electric field is given by

E = kq/r²

Where k is the coulomb constant k = 9x10⁹ Nm²/C²

q = Er²/k

q = 180,000*(0.015)²/9x10⁹

q = 4.5x10⁻⁹ C

Therefore, the charge is 4.5 nC

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Answer:

Pebble A has 1/3 the acceleration as pebble B.

Explanation:

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Same starting force, F

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3 years ago
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:

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

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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}

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Using the values for the problem:

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

This is, a percent uncertainty of 4.77 %

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Hope this gives you a little bit more information!

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