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3 years ago

How did J.J. Thomson's experiments demonstrate that Dalton's model of the atom was incorrect?

2 answers:

6 0

<span>Dalton's famous experiments using cathode ray tubes demonstrate the presence of electrons, of which Dalton was previously unaware. In his theories, Dalton posited that atoms were solid, indivisible particles. However, Thomson's experiments showed that there were in fact subatomic particles, ushering in a new era of study of the atom.</span>

morpeh [17]3 years ago

4 0

The correct option is C.
J.J Thompson demonstrated that Dalton's model of the atomic theory was wrong by showing that atoms are made up of sub particles, this is contrary to the report that Dalton gave about atoms. In his experiment, Dalton reported that atoms are indivisible, that is, they can not be broken down into smaller particles. J.J Thompson on the other was able to show through his cathode ray experiment that atoms are made up of smaller particles called electrons.

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a grape and apple and a watermelon are arranged at the corners of a triangle so that each piece of fruit is three meters away fr

777dan777 [17]

-- There are three pairs of mass with gravitational forces between them.

-- The distances between the masses are the same for each pair.

-- The only other quantity that determines the strength of the gravitational
force is the product of the masses.

-- The product of the masses is greatest for the apple and the watermelon,
so the strength of the gravitational force between them is the greatest.

5 0

3 years ago

Which class of organic molecules includes those whose main purpose is long-term energy storage?

omeli [17]

Answer:

Fats.

Explanation:

A fat molecule which is called a Triglyceride consists of two main components which are glycerol and 3 fatty acids. Fats are solids and oils are liquids at room temperature. They also include saturated and unsaturated fats.

Cells store energy for long-term use in the form of lipids called fats which provides insulation from the environment for plants and animals.

3 0

4 years ago

Air at 1.3 bar, 423 K and a velocity of 40 m/s enters a nozzle operating at steady state and expands adiabatically to the exit,

bixtya [17]

Answer:

a) T₂ = 376.905 K

b) $\eta_n$ = 95.4%

Explanation:

Given:

Initial pressure, P₁ = 1.3 bar

Initial temperature, T₁ = 423 K

Initial velocity of air, v₁ = 40 m/s

Exit pressure, P₂ = 0.85 bar

Exit velocity of the air, v₂ = 307 m/s

gas constant for air, k = 1.4

also,

the specific heat for the air, Cp = 1.005 KJ/kg.K =1.005 × 10³ J/kg.K (standard)

a) applying the energy rate balance, we have

$h_1+\frac{v_1^2}{2}=h_2+\frac{v_2^2}{2}$

where, h is the enthalpy

on rearranging we get

$\frac{v_2^2-v_1^2}{2}=h_1-h_2$

also

h = Cp × T

thus,

$\frac{v_2^2-v_1^2}{2}=C_p(T_1-T_2)$

on substituting the values, we get

$\frac{307^2-40^2}{2}=1.005\times 10^3(423-T_2)$

T₂ = 376.905 K

b) The isentropic nozzle efficiency is given as:

$\eta_n=\frac{V_2^2/2}{V_1^2/2}=\frac{Actual\ expansion}{Ideal\ Expansion}$

for an isentropic process,we have the isentropic relation as:

$\frac{T_{2s}}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}$

on substituting the values and solving, we get

$T_{2s}$ = 374.65 K

by applying the energy equation, we get the ideal exit velocity as:

$V_{2s}=\sqrt{2C_p(T_2-T_1)+V_1^2}$

on substituting the values, we get

$V_{2s}=\sqrt{2\times1.005\times10^3(423-374.65)+40^2}$

$V_{2s}=314.29\ m/s$

thus,

substituting in the formula for efficiency, we get

$\eta_n=\frac{307^2/2}{314.29^2/2}$

$\eta_n=0.954$ = 95.4%

5 0

3 years ago

What is the correct order of the planets in the solar system

Veronika [31]

That completely depends on what characteristic you use to sort them.
You could list the planets in order of their mass, diameter, density,
rotation period, surface temperature, orbital eccentricity, inclination
to the ecliptic, surface reflectivity, etc. Each sorting criterion would
give you a different list.

In the order of the length of the semi-major-axis of each planet's orbit,
the sequence, from smallest to largest, is ...

Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
the former planet Pluto.

3 0

3 years ago

Read 2 more answers

Jack (mass 55.0 kg) is sliding due east with speed 8.00 m>s on the surface of a frozen pond. he collides with jill (mass 48.0

PSYCHO15rus [73]

<span>Jill will have an angle of 36.0Â° south of east with a velocity of 5.46 m/s This problem deals with the conservation of momentum. The momentum before and after the collision will remain constant and is a vector quantity. So let's calculate the momentum before the collision. 55.0 kg * 8.00 m/s = 440 kg*m/s So we have a vector with a magnitude of 440 kg*m/s and an angle of 0Â° before the collision. After the collision, we have Jack's vector being 55.0 kg * 5.00 m/s = 275 kg*m/s So his new vector has a magnitude of 275 kg*m/s and an angle of 34.0Â° So what we have is a side angle side (SAS configuration of a triangle where one side has a length of 440, the other side has a length of 275 and the included angle is 34.0Â°. We need to calculate the length of the opposite side. Using the law of cosines, we have c^2 = a^2 + b^2 - 2ab cos(C) Substituting known values gives. c^2 = 440^2 + 275^2 - 2*440*275*cos(34) c^2 = 193600 + 75625 - 242000*0.829037573 c^2 = 269225 - 200627.0926 c^2 = 68597.90744 c = 261.9120223 So the magnitude of Jill's momentum will be 261.912 kg*m/s. And dividing by her mass, gives us a velocity of 261.912 kg*m/s / 48.0 kg = 5.4565 m/s Her angle will be south of due east. The measurement of the angle will be the same as the angle opposite to the side of length 275. And using the law of sines, we get sin(34.0)/261.9120223 = sin(X)/275 0.559192903/261.9120223 = sin(X)/275 0.002135041 = sin(X)/275 0.587136272 = sin(X) 35.95405166 = X So Jill will have an angle of 36.0Â° south of east with a velocity of 5.46 m/s</span>

5 0

4 years ago