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exis [7]
3 years ago
11

Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o

f consecutive lines is constant. Clear the equipotential lines using the Erase button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one with 12 V, one with 15 V, and one with 6 V). Don't worry about getting these exact values. You can be off by a few tenths of a volt. Which statement best describes the distribution of the equipotential lines?
1. The equipotential lines are closer together in regions where the electric field is weaker.
2. The equipotential lines are closer together in regions where the electric field is stronger.
3. The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.
Physics
1 answer:
Elza [17]3 years ago
5 0

Answer:

B or 2

Explanation:

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3 0
3 years ago
If there is a difference in air pressure between two locations, what happens?
tino4ka555 [31]

Answer:

High in the atmosphere, air pressure decreases. ... A low pressure system has lower pressure at its center than the areas around it. Winds blow towards the low pressure, and the air rises in the atmosphere where they meet. As the air rises, the water vapor within it condenses, forming clouds and often precipitation.

Explanation:

4 0
3 years ago
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The electric flux through a spherical surface is 1.4 ✕ 105 N · m2/C. What is the net charge (in C) enclosed by the surface?
Anit [1.1K]

Answer:

The  value is   Q_{net} =  1.239 *10^{-6} \  C

Explanation:

From the question we are told that

   The electric flux is \Phi =  1.4*10^{5} \  N\cdot m^2/C

     

Generally the net charge is mathematically represented as

    Q_{net} =  \Phi *  \epsilon_o

Here \epsilon_o is the permetivity of free space with value  

       \epsilon_o =  8.85*10^{-12}  \  \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So

   Q_{net} =  1.4*10^5 *  8.85*10^{-12}

=>   Q_{net} =  1.239 *10^{-6} \  C

8 0
3 years ago
he capacitor can withstand a peak voltage of 550 VV . If the voltage source operates at the resonance frequency, what maximum vo
anygoal [31]

Answer:

The maximum voltage is 39.08 V.

Explanation:

Given that,

Voltage = 550 V

Suppose, In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is 1.20\times10^{-2}\mu F

We need to calculate the resonant frequency

Using formula of resonant frequency

f=\dfrac{1}{2\pi\sqrt{LC}}

Put the value into the formula

f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}

f=2356.8\ Hz

We need to calculate the maximum current

Using formula of current

I=\dfrac{V_{c}}{X_{c}}

I=2\pi f C\times V_{c}

Put the value into the formula

I=2\pi\times2356.8\times1.20\times10^{-8}\times550

I=0.0977\ A

We need to calculate the impedance of the circuit

Using formula of impedance

Z=\sqrt{R^2+(X_{L}-X_{C})^2}

At resonant frequency , X_{L}=X_{C}

So, Z = R

We need to calculate the maximum voltage

Using formula of voltage

V=IR

Put the value into the formula

V=0.0977\times400

V=39.08\ V

Hence, The maximum voltage is 39.08 V.

4 0
3 years ago
Given a force of 100 N and an acceleration of 10 m/s^2, what is the mass?
frutty [35]

Given:

Force(F): 100 N

Acceleration: 10 m/s^2

Now we know that

F= mx a

Where F is the force acting on the object which is measured in Newton

m is the mass of the object measured in Kg

a is the acceleration measured in m/s^2

Substituting the given values in the above formula we get

100= 10m

m= 10 Kg

6 0
3 years ago
Read 2 more answers
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