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3 years ago

What was the revolution in atomic theory produced by the discovery of the electron?

2 answers:

6 0

The answer to this problem is the second item in the choices which is "The atom was comprised of particles". This atomic theory also proves that atoms cannot be destroyed nor created and it composed of many particles. It is also said that atoms of the same elements can be identical.

Orlov [11]3 years ago

3 0

Answer:

The atom was divisible.

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Solve these questions plz

8090 [49]

Answer:

For the First answer I cant answer it But I can help you :

The solid has constituent particles tightly packed and the lattice vibrations are carried out by them in their fixed position however oscillations take place about their mean position. These vibrations are increased as soon as there is increase in the temperature which eventually leads to the more chaotic motion of the constituents. At a fixed critical point of temperature, the bonds are broken and the constituent particles are spaced apart changing their phase into liquid. When more temperature is increased by gaining heat energy then the liquid changes into gas where the motion of constituent particles moving freely is dominant.

Explanation:

6 0

3 years ago

What happens to hotter air in the atmosphere?

Ber [7]

B.Warmer air molecules move faster and get farther apart. As they get farther apart, they become less dense and rise.

5 0

3 years ago

a sailboat sails 50 km to the north then 80 km in the direction 38 degrees west of south. what is the net displacement of the bo

True [87]

Answer:

35.489km

Explanation:

a diagram illustrating the question is attached

now, applying cosine rule to find the displacement c

cosine rule;

c²=a²+b²-2abCos ∅

c²=50²+80²-2(50)(80)Cos 38°

c²=2500 + 6400 - (8000×0.9951)

c²=8900-7640.589

c²=1259.411

c= $\sqrt{1259.411}$

c=35.49km

8 0

4 years ago

Read 2 more answers

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

3 years ago

Imagine a star very similar to our own Sun in both size and mass. This star moves very slightly back and forth in the sky once e

Inessa05 [86]

Answer:

(c) The planet must have a mass about the same as the mass of Jupiter,

(d) The planet must be closer to the star than Earth is to the Sun.

Explanation:

Astrometry is the ideal method to detect high-mass planets that are close to their star. That is because the gravitational effect that it will have the planet over its host star will be greater. This effect can be seen as a wobble in the star as a consequence of how they orbit a common center of mass¹. The center of mass will be closer to the most massive object, So, in the case of an extrasolar planet with masses like Jupiter (Jovian), this point will be a little bit farther from the star, making the wobble more notable than in a system with a low-mass planet.

Key terms:

Astrometry: study of the position of the stars over time in the sky.

¹Center of mass: a geometrical point in which the mass from a whole system is summed.

6 0

3 years ago