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3 years ago

13

How to calculate the unknown force

1 answer:

Lesechka [4]3 years ago

5 0

the formula to calculate the force is, F = MA

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Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T

olya-2409 [2.1K]

The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

Hope this helps!!!

4 0

3 years ago

A ball is being rolled by a normal push of 180N. It is opposed by friction which has a force of 61N and air resistance which has

Nina [5.8K]

Resultant force is basically the force left after everything is added.

if a ball is being pushed one one side with 180N, and being pushed on teh opposite side with 84N (I added friction and air resistance since they're acting on the same side), then the resultant force would be:

180N - 84N =<u> 96N</u> (you can determine whether it's positive or negative based on the direction of the vector)

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

A boy accidentally touch an electric cable and quickly withdrew his hand. Narrate the impulses involved in such action.

Lelechka [254]

Answer:

Quick action was in this as a thought of he knew he could die from this

3 0

3 years ago

Read 2 more answers

________ can occasionally cause the North Atlantic to appear green and murky.

castortr0y [4]

Answer:

algae

Explanation:

The Atlantic off the East Coast of the United States usually appears green. This is due to the presence of algae and plant life.

4 0

2 years ago