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3 years ago

10

Describe Thomson’s and Rutherford’s key discoveries about the atom.

1 answer:

muminat3 years ago

5 0

Thomson proposed the "plum pudding" model - randomly scattered positive charges and negative charges held in a sphere

Rutherford disagreed with Thomson with his gold foil experiment - proposed that at the centre of an atom was a dense, positive mass and around it were very small negative charges called electrons

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The denser a substance, the more tightly packed it is. The table below shows the volume and mass of two different substances. Co

trasher [3.6K]

Answer:

Both objects are packed equally tightly.

Explanation:

For Substance R we have

mass = 10 g

Volume = 22 cubic cm

So we know that density is defined as

$\rho = \frac{m}{V}$

here we have

$\rho = \frac{10 g}{22 cm^3}$

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For substance S we have

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So we know that density is defined as

$\rho = \frac{m}{V}$

here we have

$\rho = \frac{25 g}{55 cm^3}$

$\rho = 0.454 g/cm^3$

So here both have same density

so correct answer would be

Both objects are packed equally tightly.

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3 years ago

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A 5.40 uF parallel-plate, air capacitor has a plate separation of 3.50 mm and is charged to a potential difference of 480 V. Cal

m_a_m_a [10]

Answer:

Energy density = 0.0831 J/m³

Explanation:

Energy density of a capacitor is given by the expression

$u=\frac{1}{2}\epsilon E^2$

We have electric field ,

$E=\frac{V}{d}=\frac{480}{3.5\times 10^{-3}}=1.37\times 10^5V/m$

Here there is no dielectric

So energy density,

$u=\frac{1}{2}\epsilon_0 E^2=\frac{1}{2}\times 8.85\times 10^{-12} \times (1.37\times 10^5)^2=8.31\times 10^{-2}=0.0831 J/m^3$

Energy density = 0.0831 J/m³

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3 years ago