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3 years ago

Classical physics is a good approximation to modern physics under certain circumstances. What are they?

Physics

1 answer:

Aleks [24]3 years ago

4 0

Answer and Explanation:

In classical physics we study about macroscopic objects and in modern physics we study about microscopic objects but under some circumstances classical physics is a approximation to modern physics such as

The object must be microscopic but they should be large enough so that we can see that through a microscope
The speed of the object should be very less it should be about 1 % of the speed of the light
They should be involved with very week gravitational field

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3 years ago

Please help!! giving a lot of points

Readme [11.4K]

Question 1.

mass = 4500 kg
potential energy (p.e) = 67500 J

now, we know :

=》

$p.e = mgh$

=》

$67500 = 4500 \times 10 \times h$

=》

$67500 = 45000 \times h$

=》

$h = \dfrac{67500}{45000}$

=》

$h = 1.5 \: m$

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

mass = 4500 kg
kinetic energy = 63000 j

we know,

=》

$k.e = \dfrac{1}{2} mv {}^{2}$

=》

$63000 = \dfrac{1}{2} \times 4500 \times {v}^{2}$

=》

${v}^{2} = \dfrac{63000 \times 2}{4500}$

=》

${v}^{2} = 28$

=》

$v = \sqrt{28}$

=》

$v = 2 \sqrt{7} \: \: ms {}^{ - 1}$

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$5.29 \: \: ms {}^{ - 1}$

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3 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25

torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{V}{d}$

Put the value into the formula

$E=\dfrac{25.0}{1.50\times10^{-3}}$

$E=16666.66\ V/m$

We need to calculate the charge density

Using formula of charge density

$\sigma=E\times\epsilon_{0}$

Put the value into the formula

$\sigma=16666.66\times8.85\times10^{-12}$

$\sigma=1.474\times10^{-7}\ C/m^2$

We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}$

$C=4.484\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=4.484\times10^{-12}\times25.0$

$q=112.1\times10^{-12}\ C$

Hence, The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

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