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WITCHER [35]
2 years ago
14

Prank text my sister, I wanna see her reaction. ‪(346) 298-3870‬

Physics
1 answer:
dlinn [17]2 years ago
4 0

Answer:

she is going to be mad dude

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Asteroids do not travel in a straight line as they orbit the Sun. Per Newton's second law of motion, this implies
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A force is acting on the planet
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All circuits include
valina [46]

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a battery, wires, and a switch.

Explanation:

All circuits include?

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2 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
3 years ago
How would the weight of a 7-kilogram concrete block compared to the weight of a 7-kilogram metal sphere?
Hitman42 [59]

Answer:

C. The block and the sphere would have the same weight.

Explanation:

If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.

3 0
3 years ago
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Which term names the force every mass exerts on every other mass?
garri49 [273]
A. gravity is your answer hope this helps
3 0
3 years ago
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