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3 years ago

13

Which parts of Dalton’s atomic theory had to be revised

1 answer:

eduard3 years ago

7 0

<span> The indivisibility of an atom was proved wrong an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes </span>part<span> in chemical reactions</span>

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Amanda [17]

Answer:

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

Explanation:

The distance of the book before the lamp is moved, $d_{b} = 30 cm$

The distance of the book after the lamp is moved, $d_{a} = 90 cm$

Illumination can be given by the formula, $E = \frac{P}{4 \pi d^{2} }$

Illumination before the lamp is moved, $E_{b} = \frac{P}{4 \pi d_{b} ^{2} }$

Illumination after the lamp is moved, $E_{a} = \frac{P}{4 \pi d_{a} ^{2} }$

$\frac{E_{a}}{E_{b}} } = \frac{\frac{P}{4 \pi d_{a} ^{2} } }{\frac{P}{4 \pi d_{b} ^{2} } }$

$\frac{E_{a} }{E_{b} } = \frac{d_{b} ^{2} }{d_{a} ^{2}} \\\frac{E_{a} }{E_{b} } = \frac{30^{2} }{90 ^{2}}\\\frac{E_{a} }{E_{b} } =\frac{900 }{8100}\\\frac{E_{a} }{E_{b} } =\frac{1 }{9}$

$E_{b} = 9E_{a}$

The illumination on the book before the lamp is moved is 9 times the illumination after the lamp is moved.

6 0

3 years ago

An electron in a mercury atom changes from energy level b to a higher energy level when the

emmainna [20.7K]

The energy is 3.06 electronvolts, E = 3.06eV

1eV = 1.6 * 10^-19 J

3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J

4 0

3 years ago

Calculate the wavelength of a wave if 5 complete waves occupy a length of 20m

Nezavi [6.7K]

The answer is 20 / 5 = 4

6 0

3 years ago

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

4 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

3 years ago

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