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Scorpion4ik [409]
4 years ago
8

The primary gases of the atmosphere are _____. nitrogen and carbon dioxide nitrogen, oxygen, and argon nitrogen and oxygen nitro

gen, oxygen, and carbon dioxide
Physics
2 answers:
Paladinen [302]4 years ago
6 0
<span>The primary gases of the atmosphere are _____:
</span>
nitrogen, oxygen, and carbondioxide
inna [77]4 years ago
3 0

Answer:

nitrogen and oxygen

Explanation:

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There is a repulsive force between two charged objects when ____________ .
xxTIMURxx [149]
There is a repulsive force between two charged objects when they are of like charges/ they are likely charged (like charges repel each other)
3 0
3 years ago
Study the image of the moving car.
gayaneshka [121]

Answer:

While traveling downhill, the car’s potential is <u>increasing</u> and kinetic energy is <u>decreasing</u>

Explanation:

hope this helps!

6 0
2 years ago
Read 2 more answers
Examine the scenario.
vovangra [49]

Answer:

acceleration 8 km/h/s south

Explanation:

First of all, let's remind that a vector quantity is a quantity which has both a magnitude and a direction.

Based on this definition, we can already rule out the following two choices:

distance: 40 km

speed: 40 km/h

Since they only have magnitude, they are not vectors.

Then, the following option:

velocity: 5 km/h north

is wrong, because the car is moving south, not north.

So, the correct choice is

acceleration 8 km/h/s south

In fact, the acceleration can be calculated as

a=\frac{v-u}{t}

where

v = 40 km/h is the final velocity

u = 0 is the initial velocity

t = 5 s is the time

Substituting,

a=\frac{40 km/h-0}{5 s}=8 km/h/s

And since the sign is positive, the direction is the same as the velocity (south).

7 0
4 years ago
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

7 0
3 years ago
What crisis is occurring in California?
nlexa [21]
Hey there!

Option A

A climate crisis is occurring in California where sudden forest fires occur due to this .
3 0
2 years ago
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