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kkurt [141]
2 years ago
9

By moving a 20 nC charge from point A to point B, you determine that the electric potential at B is 100 V . Part A What would be

the potential at B if a 40 nC charge were moved from A to B
Physics
1 answer:
ivolga24 [154]2 years ago
4 0

Answer:

Potential at B would be 100V

Explanation:

The electric potential is defined as the work done to bring a unit positive charge from infinity to some point in the field.

We always determine the potential with respect to some reference point. Let the potential at A be zero. If the potential at B is V, then work done to bring charge q from A to B = qV

which is the electric potential energy.

If instead we use some charge Q, the electric potential <em>energy</em> will be QV, but the electric potential will always be V.

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Sometimes the tar of a newly paved road gets sticky on a very hot summer day. What causes this?
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I’m pretty sure it is caused by the heat of the sun warming it up back into its original state of tar
6 0
3 years ago
A person walks due south from point A for 500 yards and then due west for 300 yards, arriving at point B. Answer the following q
Pepsi [2]

The total displacement of the person walking from point A to point B is 300 yards.

As shown in the figure we can conclude that the required method to calculate the total displacement is the Pythagoras theorem.

<h3>Pythagoras theorem in brief :</h3>

According to the Pythagorean Theorem, the square that represents the hypotenuse, or side of a right triangle that faces the right angle, is equal to the total of the squares on the triangle's legs.(or, in popular algebraic notation, a^2 + b^2 = c^2).

<h3>Calculation: </h3>

Let,

a = 500

b=  300

Hence by using Pythagoras' theorem

Total displacement of the person = \sqrt{500^{2}  + 300^{2} } = \sqrt{900000} = 300

Thus the total displacement of the person from starting point is 300 yards.

Learn more about the displacement examples here:

brainly.com/question/11188852

#SPJ4

5 0
2 years ago
A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm
Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

   v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

          λ=2 L

          λ=2 x 0.577 = 1.154 m

we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

             f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0
3 years ago
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