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Afina-wow [57]
3 years ago
6

Two charges, -20 C and +4.7 C, are fixed in place and separated by 3.0 m. a. At what spot along a line through the charges is th

e net electric field zero? Give the distance of the spot to the positive charge in meters (m). (Hint: The spot does not necessarily lie between the two charges.) b. What would be the force on a charge of +14 C placed at this spot?

Physics
1 answer:
Pani-rosa [81]3 years ago
6 0

The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.

Explanation:

Let at A both net electric field is zero then

At A ,E1=E2

E1=k*Iq1I / (d+x)^2

E2=k*Iq2I /x^2

Equating both

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A Ray of light falling on rough surface follows the laws of reflection but no image of the object placed before it is C explain
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Answer:

The light rays falling on a rough surface does follow the laws of reflection. The light rays are incident parallel on the rough surface but due to uneven surface the light rays are not reflected parallel rather they are reflected in different direction. Hence, no image is formed.

4 0
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A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it
Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

                                                     \Phi=B\ A\ cos\theta

We are told the values of \Phi and B, then we can calculate the magnitude

                                                      \frac{\Phi}{A}=B\ cos\theta

3 0
3 years ago
iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms
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Answer:

8.60 g/cm³

Explanation:

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Therefore:

\frac{2}{a^{3} } = \frac{N_{A} * p }{A}

This implies that:

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Calculate the average orbital speed of Ceres in
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8 0
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a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

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In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

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In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
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