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emmasim [6.3K]
3 years ago
6

When electrons are removed from the outermost shell of a calcium atom, the atom becomesA. an anion that has a larger radius than

the atomB. an anion that has a smaller radius than the atom.C. a cation that has a larger radius than the atom.D. a cation that has a smaller radius than the atom.
Physics
2 answers:
hoa [83]3 years ago
6 0

Answer:

D: a cation that has a smaller radius than the atom

Explanation:

Got it right on edg :)

Free_Kalibri [48]3 years ago
3 0

Answer:

D. a cation that has a smaller radius than the atom.

Explanation:

When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.

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Explanation:

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A professor drives off with his car (mass 870 kg), but forgot to take his coffee mug (mass 0.47 kg) off the roof. The coefficien
SSSSS [86.1K]

Answer:6.86 m/s^2

Explanation:

Given

mas of car=870 kg

coffee mug mass=0.47 kg

coefficient of static friction between mug and roof \mu _s= 0.7

Coefficient of kinetic Friction \mu _k=0.5

maximum car acceleration is \mu \times g

here coefficient of static friction comes in to action because mug is placed over car . If mug is moving relative to car then \mu _k is come into effect

a_{max}=0.7\times 9.8=6.86 m/s^2

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3 years ago
Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho
uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

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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:

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

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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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A force F is exerted at a distance D from the axis of rotation of a wrench and is exerted at an angle θ to the wrench (θ is the
Katyanochek1 [597]

Answer:

Torque = R X F = R * F sin theta

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-f * 2 = F sin theta

f = -F sin theta / 2

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3 years ago
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