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ICE Princess25 [194]
3 years ago
9

Which value is equivalent to 178 centimeters?

Physics
1 answer:
tester [92]3 years ago
7 0

178 centimeters is exactly, precisely, the same length as ...

-- 0.00178 kilometer
-- 1.78 meters
-- 1,780 millimeters
-- 1,780,000 micrometers
-- 1,780,000,000 nanometers

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A point charge is placed at the center of a spherical Gaussian surface. The electricflux ΦEischangedif(a) a second point charge
Simora [160]

Answer:

(b) the point charge is moved outside the sphere

Explanation:

Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.

\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.

If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.

Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.

7 0
3 years ago
1. The speaker uses needles from two
GuDViN [60]

Answer:

1. Georgia

2. Ten

3. Repeating

Explanation:

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6 0
4 years ago
If you apply 100.0 N of force to lift an object with a single, fixed pulley, then what is the resistive force?
quester [9]
<span>If you apply 100.0 N of force to lift an object with a single, fixed pulley, then the resistive force is also equivalent to 100 Newtons of force. Since the weight of the object was not mentioned, it is assumed that it has already been taken into account in the 100 N value of force. This follows Newton's law of motion of equal action and reaction.</span>
6 0
3 years ago
how does electromagnetism effect the function of a generator (using magnetism to produce electrical energy)
Sati [7]

Explanation:

We know that a changing magnetic field induces a current in a conductor. For that reason a generator basically consist an element that produces a magnetic field that changes over time and a conductor where the current will be induced.

This element that produces a magnetic field can be one of the following:

- A permanent magnet: Which is basically like a regular magnet. The magnetic field that a permanent magnet produces does not change over time, we need a motor or any other external force to move the axis of the generator and cause the magnetic field to change.

- An electro-magnet. Which is basically a DC current flowing through a conductor. Basically, when current flows through a conductor it behaves exactly like a magnet. So what we commonly do, is to connect a conductor to a DC battery, and it will create a magnetic field.

Like we are using a DC battery to create a magnetic field, then the magnetic field won't change over time either. So we still need an external force to move the axis of the generator to produce AC electricity.

4 0
3 years ago
Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.
8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

The distance from the center of the square to one of the corners is \sqrt2 L/2 = 0.035m

V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

r_1 = 0.05\sqrt2m\\r_2 = 0.05m

V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between  energies. This is the work-energy theorem. So,[tex]W = U_b - U_a

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}

4 0
3 years ago
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