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denpristay [2]
4 years ago
6

Which formula represents Gay-Lussac's law?

Physics
1 answer:
lana66690 [7]4 years ago
6 0
The formula of Gay-Lussac's law is Pi/Ti = Pf/T<span>f</span>
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Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value o
Anna35 [415]

Answer:

the correct one is 2. the equipotential lines must be closer together where the field has more intensity

Explanation:

The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.

In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by

           V = k q / r

also the electric field and the electuary potential are related

           E =  - \frac{dV}{dr}

therefore the equipotential lines must be closer together where the field has more intensity

When checking the answers, the correct one is 2

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3 years ago
The Moon is attracted to the Earth. The mass of the Earth is 6.0x1024 kg and the mass of the Moon is 7.4x1022 kg. If the Earth a
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Sorry I don't know the answer

7 0
3 years ago
a wave's amplitude is 0.5 meters. if its amplitude is increased to 1 meter, how much does its energy change?​
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The energy becomes 4 times greater.
5 0
3 years ago
What is the initial position of the object?
Mariana [72]

Answer:

e:10 m   b: -5 m/s b.The object decreases its velocity.

Explanation:

7 0
3 years ago
The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in
chubhunter [2.5K]

Answer:

The angular separation equals 0.35^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})

\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})

Using this equation for both the colors separately we have

\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}

Similarly for violet light we have

\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}

Thus the angular separation becomes

\Delta \theta =14.48-14.13=0.35^{o}

6 0
4 years ago
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