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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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vitfil [10]
B is the answer. because it is a physical change

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3 years ago
a 0.25 kg arrow is moving at 5 m/s and hits a 0.10 kg apple. The apple sticks to the arrow, and both move to the right together.
MatroZZZ [7]

Answer:

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

Explanation:

Use the law of conservation of momentum to solve this problem. In this case the law can be written as follows:

m_{arrow}v_{arrow}+m_{apple}\cdot 0= (m_{arrow}+m_{apple})v_{both}

from which the desired velocity can be isolated:

\frac{m_{arrow}v_{arrow}}{m_{arrow}+m_{apple}}= v_{both}\\v_{both} = \frac{0.25kg\cdot 5\frac{m}{s}}{0.35kg}=3.6\frac{m}{s}

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

3 0
3 years ago
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7 0
3 years ago
A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one
allochka39001 [22]

Answer:

175.3 N

Explanation:

The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.

There are two forces acting on the ball at the lowest point of motion:

- The tension in the string, T , upward

- The weight of the ball, mg, downward

The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:

T-mg=m\frac{v^2}{r}

where the term on the right is the expression for the centripetal force, and where:

T is the tension in the string

mg=100 N is the weight of the ball

m=\frac{mg}{g}=\frac{100}{9.8}=10.2 kg is the mass of the ball

v = 5.7 m/s is the speed of the ball at the lowest point

r = 4.4 m is the length of the rope, so the radius of the circle

Solving for T, we find the tension in the string:

T=mg+m\frac{v^2}{r}=(100)+(10.2)\frac{5.7^2}{4.4}=175.3 N

7 0
3 years ago
Express force in terms of base units​
MariettaO [177]
<h2><u>Answer:</u></h2>

<u></u>

<u>Force = kgm/s² ⟶ Newton</u>

<h2><u>Explanation:</u></h2>

<u></u>

According to the formula we've learnt,

  • Force = mass × acceleration.

To find force in terms of base units, we must first identify the base SI units of mass & acceleration.

  • Base SI unit of mass = kg (kilogram)
  • Base SI unit of acceleration = m/s² (meter per second squared)

So, the base unit of force is,

  • Force = mass × acceleration.
  • Force = kg × m/s²
  • <u>Force = kgm/s² ⟶ Newton</u>

________________

Hope it helps!

\mathfrak{Lucazz}

4 0
2 years ago
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