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iren2701 [21]
2 years ago
10

When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoin

t of the hypotenuse is 20 V. When the charge qa is removed, the potential at the midpoint becomes 15 V. When, instead, the charge qb is removed (qa and qc both in place), the potential at the midpoint becomes 12 V. What is the potential at the midpoint if only the charge qc is removed from the array of charges?
Physics
1 answer:
earnstyle [38]2 years ago
7 0

Answer:

8v

Explanation:

First we apply super position principle

Vt= v1 + v2+ v3

Remove qa

But vt= 20v

So V = v2+v3

V1= 20-15

= 5v

Remove qb

V= v1+v3

V=8v

So the potential when qa and qc are remove is the potential due to qb

Which is 8v

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remains the same, but the apparent brightness is decreased by a factor of four.

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2 years ago
What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u
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Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 ×  10^{14} Hz

solution

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E=h\nu_{o}    ..................1

here E is the energy of electron per atom E=\frac{x}{N}  and h is plank constant i.e. 6.626\times10^{-34} Js  and x is  binding energy

and here N is the Avogadro constant = 6.023\times10^{23}

so E will \frac{x}{6.023\times10^{23}}  

E = 3.19\times10^{-19}  J  

so put value in equation 1 we get

\frac{x}{6.023\times10^{23}} = 2.50 ×  10^{14} × 6.626\times10^{-34} Js  

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x = 99770.99

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