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Alla [95]
2 years ago
8

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

f the pair of charges is +5.4×10−8J. When the second charge is moved to point b, the electric force on the charge does −1.9×10−8J of work.
What is the electric potential energy of the pair of charges when the second charge is at point b?
Physics
1 answer:
Brrunno [24]2 years ago
7 0

The electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

<h3>Electric potential energy</h3>

When work is done on a positive test charge to move it from one location to another, potential energy increases and electric potential increases.

The electric potential energy between the charges when the second charge is at point b is calculated as follows;

ΔU = -w

Ui - Uf = w

Uf = Ui - w

where;

Uf is the final potential energy

Ui is the initial potential energy

w is the work done by the force

Uf = 5.4 x 10⁻⁸ J - (-1.9 x 10⁻⁸J)

Uf = 5.4 x 10⁻⁸ J + 1.9 x 10⁻⁸ J

Uf = 7.3 x 10⁻⁸ J

Thus, the electric potential energy of the pair of charges when the second charge is at point b is 7.3 x 10⁻⁸ J.

Learn more about electric potential energy here: brainly.com/question/14306881

#SPJ1

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Answer:

32 pounds

Explanation:

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5 0
3 years ago
Which object has more gravitational potential energy? Use PE = m × g × h, where g = 9.8 meters/second2.
Andrew [12]
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6 0
3 years ago
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With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as
Nonamiya [84]

Answer:

|Δf| = 37.3 kHz

Explanation:

given,

peak velocity = 4 m/s

speed of the sound = 1500 m/s

frequency = 7 MHz

v = C\dfrac{\pm \dlta f}{2 f_0}

\delta f = \pm 2 f_0 (\dfrac{V}{C})

\delta f = \pm 2\times 7 (\dfrac{4}{1500})

           =\pm 0.0373 MHz

           = 37.3 kHz

|Δf| = 37.3 kHz

hence, frequency shift between the opening and closing valve is 37.3 kHz

4 0
3 years ago
Desperado, a roller coaster built in Nevada, has a mass of 800 kg. It also has a vertical drop of 225 feet down the first hill.
weeeeeb [17]

Answer:

the work done by friction on the car is 524,582 J.

Explanation:

Given;

mass of the roller coaster, m = 800 kg

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The time taken for the coaster to drop down the hill is calculated as;

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W =F\ \times \ d = \frac{mv}{t} \times \ d\\\\W = \frac{800 \ \times \ 35.76  }{3.74} \times \ 68.58\\\\W = 524,582 \ J

Therefore, the work done by friction on the car is 524,582 J.

6 0
3 years ago
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olga55 [171]
The scientist is likely to be studying kinematics.
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7 0
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