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3 years ago

I need help if you can help me to get 10 points.

Physics

2 answers:

Alinara [238K]3 years ago

6 0

Answer:

It allows the reader to feel more connected to her purpose

Explanation:

If Cooper is using personal examples for her argument, she is allowing the reader see more of her point of view, which can also lead them to feel connected to her purpose

sveticcg [70]3 years ago

3 0

Answer: Is B

Hope this helps :)

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How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position

e-lub [12.9K]

The answer is C. Final position minus initial position.

5 0

3 years ago

explain how many minimum number of geostationary satellites are required for global coverage of T.V transmission

kobusy [5.1K]

Answer:I honestly don't know

Explanation:

4 0

3 years ago

Suppose a constant net force of 345 N is applied to slide a heavy stationary couch across the

ira [324]

Answer:

517.5Ns

Explanation:

F=(MV - MU)/t

where MV - MU is the change in momentum,

therefore, MV - MU = Ft

= 345 X 1.

= 517.5Ns

4 0

3 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

A proton, an electron, and a helium nucleus all move at speed v . Rank their de Broglie wavelengths from largest to smallest.

Varvara68 [4.7K]

Ranking of de Broglie wavelengths from largest to smallest is electron > proton > helium

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .....(1) , where λ is known as de Broglie wavelength and m is mass , v is velocity , h = Plank’s constant.

From equation (1) wavelength and mass has an inverse relation .

Mass of helium is 4 times the mass of the proton and proton has a greater mass than electron.

According to equation (1) , less the mass higher will be the wavelength

Hence electron having less mass have higher wavelength and then proton and then helium having large mass will have less wavelength .

Thus, order should be electron > proton > helium .

Learn about de brogile wavelength more here :

brainly.com/question/16595523

#SPJ4

8 0

1 year ago