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3 years ago

6. In a Circular Coil the magnetic field lines take which shape?

Physics

1 answer:

Alina [70]3 years ago

3 0

Answer is concentric circles

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An object moves 2.5 m. This is an example of a _______. Question 3 options: direction distance velocity speed

Hitman42 [59]

Distance, since distance represents how far something has travelled, which would be in our case 2.5m.

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3 years ago

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Why is coal considered a nonrenewable resource? Check all that apply.

sveta [45]

Answer:

cola forms so slowly that we could use up the supply that exists now

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3 years ago

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The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particl

Lyrx [107]

Answer:

a) F = 21.16 N, b) a = 3.17 10²⁸ m / s

Explanation:

a) The outside between the alpha particles is the electric force, given by Coulomb's law

F = $k \frac{ q_1 q_2}{r^2}$

in that case the two charges are of equal magnitude

q₁ = q₂ = 2q

let's calculate

F = $9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }$

F = 21.16 N

this force is repulsive because the charges are of the same sign

b) what is the initial acceleration

F = ma

a = F / m

a = $\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }$ 21.16 / 4.0025 1.67 10-27

a = 3.17 10²⁸ m / s

this acceleration is in the direction of moving away the alpha particles

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3 years ago

To test your FHE brother's claims about his car, you accept his offer of a ride about campus. Not wanting to exceed the speed li

damaskus [11]

Answer:

4.78 m/s^2

Explanation:

A)From he FBD

we can write

Tcosθ= mg

Tsinθ= ma

dividing we get tanθ= a/g

a= g tanθ

a= 9.81 tan26°= 4.78 m/s^2

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4 years ago

At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this

Anastasy [175]

Answer:

Part A:

$\rm 2.8\times 10^3\ Volts.$

Part B:

$\rm 5.6\times 10^{-5}\ J.$

Explanation:

Part A:

Potential energy of charge at the given point, $\rm U=42\ \mu J=42\times 10^{-6}\ J.$

Charge, $\rm q=15\ nC = 15\times 10^{-9}\ C.$

The potential energy at a point due to a charge is defined as

$\rm U=qV$ .

where,

V = electric potential at that point.

Therefore,

$\rm V=\dfrac{U}{q}=\dfrac{42\times 10^{-6}}{15\times 10^{-9}}=2.8\times 10^3\ Volts.$

Part B:

Now, if the charge at that point is replaced with $\rm q_1 = 20\ nC = 20\times 10^{-9}\ C.$ , then the electric potential energy at that point is given by

$\rm U=q_1V = 20\times 10^{-9}\times 2.8\times 10^3=5.6\times 10^{-5}\ J.$

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3 years ago