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Vladimir79 [104]

3 years ago

15

How does potential difference behave in a parallel circuit

2 answers:

Illusion [34]3 years ago

6 0

The same voltage will appear across all resistors in parallel.

Fudgin [204]3 years ago

3 0

In Parallel circuits, the electric potential difference across each resistor is the same. ... In a parallel circuit, the voltage drops across each of the branches is the same as the voltage gain in the battery. Thus, the voltage drop is the same across each of these resistors. happy to help:)

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A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = $\frac{\sqrt{3} }{4} a^{2}$

= $\frac{\sqrt{3} }{4} (0.25)^{2}$

= 0.0271 m³

Lets find electric flux,

Electric Flux = E. A

= 260×0.0271

= 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

Electric flux through three sides = $\frac{7.046}{3}$

= 2.348 N m²/C

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A dog runs from point O to B. The dog reaches A at 2.3 s into the sprint. Then the dog reaches C at 4.1 s. What is the average v

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An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve

Vlad1618 [11]

Answer:

$v=0.9833\ c$

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side, b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 $kg/m^3$

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184= \sqrt{1-\frac{v^2}{c^2}}$

$0.033= 1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

$v=0.9833\ c$

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

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3 years ago