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Arte-miy333 [17]
2 years ago
6

A positive charge is moved from one point to another point along an equipotential surface. The work required to move the charge:

______.
a. is positive.
b. is zero.
c. is negative.
d. depends on the magnitude of the potential.
e. depends on the sign of the potential.
Physics
1 answer:
liberstina [14]2 years ago
7 0

Answer:

b . is zero.

Explanation:

An Equipotential surface is the one that has electric potential same throughout the surface. The formula for work done in an electric field is given by

dW= qdV

Now, in an Equipotential surface dV= 0.

Hence the work done in moving a charge from one place to the other in an electric field is zero.

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Which value is NOT equivalent to the others?
pishuonlain [190]

Answer:

A) 5 × 1011nm

I think It's A

8 0
2 years ago
A ball is thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy
natka813 [3]

Answer:

Option A

Air resistance

Explanation:

Despite the law of conservation of energy stating that energy can neither be created nor destroyed but can only be transformed from one state to the other, the process of transformation is not 100% efficient since some energy losses occur due to friction and air resistance. Therefore, when the final energy is slightly less than the original dor this case, it's due to energy loss due to air resistance.

8 0
3 years ago
A positive charge Q1 = 30.5 nC is located at the origin. A negative charge Q2 = -3.5 nC is located on the positive x-axis p = 3
Dmitriy789 [7]

Answer:

The points where the electric field due to the two charges are 0 include

x = 4.54 cm and x = 2.24 cm.

Explanation:

For a positive charge, the electric field is directed outward from the charge,

For a negative charge, it is directed towards it.

Since the positive charge is at the origin, the, and the negative charge is at p = 3cm on the x-axis, let the point where the electric field is 0 be x.

For ease of calculation, we will be looking for x, along the positive x-axis in between the two charges.

Electric field at a point due to a particular charge is given as

E = (kq/r²)

k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²

r = distance of the point from the charge.

q = the charge

q₁ = 30.5 nC = (30.5 × 10⁻⁹) C

q₂ = -3.5 nC = (3.5 × 10⁻⁹) C

For the positive charge

E₁ = (kq₁)/x²

= (9.0 × 10⁹ × 30.5 × 10⁻⁹)/x²

E₁ = (274.5/x²) (eqn 1)

For the negative charge

E₂ = [kq₂/(x-0.03)²]

E₂ = 31.5/(x-0.03)²

E₁ + E₂ = 0

(274.5/x²) - [31.5/(x-0.03)²] = 0 (with the assumption that the point is to the right of both charges)

(274.5/x²) =-[31.5/(x-0.03)²]

274.5 (x-0.03)² = 31.5x²

274.5 (x² - 0.06x + 0.0009) = 31.5x²

274.5x² - 16.47x + 0.24705 = 31.5x²

243x² - 16.47x + 0.24705 = 0

Solving this equation

x = 0.0454 m or x = 0.0224 m

So, the points where the electric field due to the two charges are 0 include

x = 4.54 cm and x = 2.24 cm.

Hope this Helps!!!

7 0
3 years ago
Read 2 more answers
Seeing how strong our gravitational pull is here on Earth, would it be possible to kill someone if you drop a penny off the Empi
nadezda [96]
That's an urban legend that's been around for a long time.
It was tested on "Mythbusters".  The conclusion was that
it would sting pretty good, and maybe raise a lump, but
it would never penetrate the skull, and it couldn't kill.
4 0
3 years ago
A 0.0427 kg racquet-ball is moving
Gwar [14]

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball (m_1) = 0.00427 kg

Velocity of racket-ball before collision (v_{1i}) = 22.3 m/s

Velocity of racket-ball after collision with box (v_{1f}) = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision v_{2i} = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision v_{2f}= 1.53 m/s

To find the mas of the box m_2.

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

p_i=p_f

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

We can plugin the given value to find m_2

(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)

0.9522+0=-0.4911+1.53m_2

Adding both sides by 0.4911

0.9522+0.4911=-0.4911+0.4911+1.53m_2

1.4433=1.53m_2

Dividing both sides by 1.53.

\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}

0.9433=m_2

∴ m_2=0.9433 kg

Mass of the box = 0.9433 kg (Answer)

4 0
2 years ago
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