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1 year ago

When the leveling bulb is higher than the water level, the pressure in the system is greater than atmospheric pressure.

1 answer:

4 0

When the leveling bulb is higher than the water level, the pressure in the system is greater than atmospheric pressure. This statement is true.

In the physical sciences, pressure is defined as the stress at a point within a confined fluid or the perpendicular force per unit area. A 42-pound box with a bottom area of 84 square inches will impose pressure on a surface equal to the force divided by the area it is applied to, or half a pound per square inch.

Atmospheric pressure, which is roughly 15 pounds per square inch at sea level, is the weight of the atmosphere pressing down on each unit area of the Earth's surface. Pascals are used to express pressure in SI units; one pascal is equivalent to one newton per square meter. Almost 100,000 pascals of atmospheric pressure are present.

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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

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3 years ago

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r <

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Answer:

Hence the answer is E inside $= KQr_{1} /R^{3}$ .

Explanation:

E inside $= KQr_{1} /R^{3}$

so if r1 will be the same then

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so if R become 2R

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2 years ago

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Answer:

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Explanation:

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