In the diagram, q1 = +4.88*10^-8 C.
What is the potential difference when
you go from point A to point B?
Include the correct sign, + or - .
B
0.538 m
1.36 m
91 |
(Hint: Does V go up or down when you go
from B to A?) (Unit = V)

Answers

Answer 1

Answer:

ΔV = 1139.3 V = 1.139 KV (+ve sign shows V goes up)

Explanation:

The potential difference while moving from point A to Point B is given as follows:

[tex]\Delta V = V_B-V_A[/tex]

where,

ΔV = potential difference from A to B = ?

[tex]V_A[/tex] = Potential at point A = [tex]\frac{kq}{r_A}[/tex]

[tex]V_B[/tex] = Potential at point B = [tex]\frac{kq}{r_B}[/tex]

Therefore,

[tex]\Delta V = \frac{kq}{r_B}-\frac{kq}{r_A}\\\\\Delta V = kq(\frac{1}{r_B}-\frac{1}{r_A})[/tex]

where,

k = Colomb's Constant = 9 x 10⁹ N.m²/C²

q = magnitude of charge = 4.88 x 10⁻⁸ C

[tex]r_A[/tex] = distance of point A from charge = 1.36 m

[tex]r_B[/tex] = distance of point B from charge = 0.538 m

Therefore,

[tex]\Delta V = (9\ x\ 10^9\ N.m^2/C^2)(4.88\ x\ 10^{-8}\ C)(\frac{1}{0.538\ m}-\frac{1}{1.36\ m})\\\\\Delta V = (439.2 N.m^2/C)(2.59\ /m)[/tex]

ΔV = 1139.3 V = 1.139 KV (+ve sign shows V goes up)

Answer 2

The answer is 492.87.

Correct on Acellus


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Answers

Answer:

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Answers

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Answers

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I took the quiz

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Answers

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Answers

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Answers

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Answers

1:-

[tex]\\ \sf\longmapsto cos\Theta=\dfrac{B}{H}[/tex]

[tex]\\ \sf\longmapsto cos\Theta=\dfrac{4}{5}[/tex]

2:-

[tex]\\ \sf\longmapsto tan\Theta=\dfrac{P}{B}[/tex]

[tex]\\ \sf\longmapsto Tan\Theta=\dfrac{3}{4}[/tex]

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P denotes to Perpendicular.B denotes to Base.H denotes to Hypotenuse.

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Cos A = 4/5

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Explanation:

Cos A = Adjacent/Hypotenuse = 4/5

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Answers

Answer:

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