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ivolga24 [154]
3 years ago
10

A body of mass 10 kg is moving at a constant speed of 20 m s over a frictionless surface the work done by the weight is

Physics
1 answer:
UNO [17]3 years ago
8 0

Answer:

200j

Work Done=Wieght×Speed

10×20

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Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-
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Answer:

C-14 has two more neutrons.

Explanation:

4 0
3 years ago
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Radioactive isotopes of an atom aren't as stable as you would think, so the correct answer is: C. Less Stable

(I can verify this because I took the test.)

Hope this helps! Have a wonderfully wonderful day!

Cheers mate!

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If the valence electrons were removed, what would be the ion charge of the element?
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If valence electron i.e electron from 3s orbital is removes, sodium will have configuration same as Neon with a +1 charge on it.
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