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olya-2409 [2.1K]
3 years ago
12

Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the tot

al force exerted by the four charges on a charge Q located a distance b along a line perpendicular to the plane of the square and equidistant from the four charges?
Physics
1 answer:
guapka [62]3 years ago
3 0

Answer:

F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

R = \frac{d}{\sqrt2}

now we have

E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}

so the force on the charge is given as

F = QE

F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}

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Art [367]

Answer:

In physics, power is the amount of energy transferred or converted per unit time. In the ... Power (physics) ... Angular acceleration / displacement / frequency / velocity. show. Scientists ... Hence the formula is valid for any general situation. ... because they define the maximum performance of a device in terms of velocity ratios

Explanation:

8 0
2 years ago
Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
2 years ago
A water wave has a speed of 23.0 meters/second. If the wave’s frequency is 0.0680 hertz, what is the wavelength?
tankabanditka [31]
     Using the Fundamental Equation of Wave, we have:

v=\lambda f \\ \lambda= \frac{23}{0.068}  \\ \boxed {\lambda \approx 338m}

Letter B
3 0
3 years ago
Read 2 more answers
At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh
Afina-wow [57]

Answer:

W = 1.22 \times 10^9 J

Explanation:

Initial potential energy of the given spacecraft is given as

U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}

so we have

U = - \frac{Gm}{r}(M_e + M_m)

so we have

M_e = 5.98 \times 10^{24} kg

M_m = 7.35 \times 10^{22} kg

m = 1160 kg

r = 3.84 \times 10^8 m

U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})

U = -1.22 \times 10^9 J

now total work done to move it to infinite is given

W = 0 - U

W = 1.22 \times 10^9 J

6 0
2 years ago
WHOEVER GET IT RIGHT GETS 50 POINTS
AVprozaik [17]

Answer:

ANSWER BELOW I

                             I

                            V

Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in

m/s2

. For example, for Earth, 445 N = 45.4 × 9.8

m/s2

:Notice that the x-axis values will be gravity in

m/s2

, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).

3 0
3 years ago
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