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2 years ago

A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm

Physics

1 answer:

Effectus [21]2 years ago

8 0

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

$V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm$

So, the thickness of the film is equal to 3.1 cm.

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3 years ago

How is the temperature of a gas related to the kinetic energy of its particles?

devlian [24]

Answer:

As the temperature increases, the kinetic energy of the particles increases.

Explanation:

When the temperature of the substance increases, the velocity increases which makes the movement of the particles to speed up. This causes the particles to increase. Therefore, as the temperature increases, the kinetic energy of the particles also increases.

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3 years ago

Read 2 more answers

Earth's core is the source of the energy that drives the movement of tectonic

Ostrovityanka [42]

Answer:

B. Convection

D. Conduction

Explanation:

Conduction and convection are the two most prominent processes that helps transfer energy outward to the earth's crust.

Energy within the core is a function of the radioactive decay and frictional heating.
Also, heat that accreted during the formation of the earth is a significant source of internal energy.
The heat is conducted away by the process of convection. This is possible due to temperature differences between different parts of the earth
Conduction is made made possible due to the metallic bodies in the core and other part of the inner earth.

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2 years ago

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

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3 years ago

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butalik [34]

Answer:

yes \\

have a nice day T_T

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3 years ago