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NemiM [27]
3 years ago
12

Earth is the third planet from the Sun. This placement most affects Earth's unique ____ and ____ conditions.

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0
1 is  A
2 is  D
I believe this is the answer <span />
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(6.MS-ETS2-1(MA).) The electrons in __________ move about more freely than the electrons in insulators which is why this type of
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Answer:

A) conductors

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A conductor can be defined as any material or object that allows the free flow of current or electrons (charge) in one or more directions in an electrical circuit. Some examples of a conductor are metals, tungsten, copper, aluminum, iron, graphite, etc.

Basically, the main purpose of a conductor in physics is to provide a low-resistance path between electrical circuits or components. This low-resistance path is to ensure that the electrical components allows the free flow of electrons and thus, enabling charge transfer.

Hence, the electrons in conductors move about more freely than the electrons in insulators which is why this type of material can be used to create electric circuits because it would significantly provide a low-resistance path between the electric circuits.

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Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The forc
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3 years ago
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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
4 years ago
Does current in Helmholtz coils flow in the same or opposite direction through each coil? Explain.
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To get a uniform field in the central region between the coils, current flows in the same direction in each. 
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