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horsena [70]
3 years ago
13

The mass of a substance was found to be 8.2 g. By using the water displacement method the volume was found to be 3.6 cm3. Calcul

ate the density
Physics
2 answers:
Tatiana [17]3 years ago
6 0

Answer:

D= 2.3g/ml

Explanation:

Vikki [24]3 years ago
5 0
D= M/V

D=8.2/3.6

D= 2.3g/ml
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Early in the twentieth century, a group of German psychologists noticed that people tend to organize a cluster of sensations int
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Answer:

The correct option is D

D) Gestalt

Explanation:

In the earlier 20th century, a group of German scientists named Max Wertheimer, Wolfgang Köhler and Kurt Koffka, noticed that people tend to organzie a cluster of information into a Gestalt.

Gestalt is a German word which means the way things are placed or put together as a whole. In the field of phychology, it can be defined as a pattern or configuration. In this context, it included the human mind and its behavior as a whole.

The Gestalt theory states that:

"The whole of anything is greater than its parts and that attributes of the whole can't be deduced by analyzing any of the parts on their own accord"

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What happens if you drop a chicken wing from the top of the Eiffel Tower
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It either goes WEEEEEEE. Or it just breaks apart.

Explanation:

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What is the answer to this?
jeka57 [31]

Answer:

I think it is the forth one

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2 years ago
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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

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