All categories

3 years ago

The moon is closest to earth at ____.

2 answers:

8 0

The point of the orbit closest to Earth<span> is called perigee, while the point farthest from </span>Earth<span> is known as apogee</span>

Fynjy0 [20]3 years ago

7 0

Answer:

The answer is Perigee

Explanation:

You might be interested in

2. What biotic and abiotic factors might influence the wolf and moose population numbers? List 1 biotic factors 1 abiotic factor

Ulleksa [173]

Answer:

Abiotic - sun Biotic- Plants

Water supply, climate, shape of the land, vegetation, soils and availability of natural resources.

Explanation:

Abiotic means non living so the sun is non living. The sun gives us warmth and the ability to survive

Biotic means alive or living- plants are living and give off oxygen and take in carbon to help us live umm ok

6 0

2 years ago

The process of burning fuel is called?

user100 [1]

The process of burning fuel is Combustion

5 0

3 years ago

Read 2 more answers

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

While running around the track at school, Milt notices that he runs due East on the 100m homestretch and due West on the 100m ba

Yuri [45]

Answer a would be correct since velocity is a vector and has a magnitude and a direction. In this case v₁ = - v₂.

8 0

3 years ago

Read 2 more answers

Elanso [62]

Answer:

Well, their are two answers in their. It would be Ask their Parent for assistance in persuading and Ask for an opportunity to earn extra credit:)

Explanation:

3 0

3 years ago

Read 2 more answers