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A disk made of many small particles of Rock and ice in orbit around a planet is called

aleksandrvk [35]

The correct answer is ring. A disk made of many small particles of rock and ice in orbit around a planet is called ring. The rings of the planet Saturn is considered to be the most extensive planetary rings system in the Solar System.

3 0

3 years ago

A construction worker drives to his work 43 km away at an average speed of 62 km/h. Before he gets out of his car he realizes he

swat32

Answer:

Explanation:

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7 0

3 years ago

PLEASE HELPPPPPPPPPPPPPP

MrRa [10]

Your answer is c hope I helped

7 0

3 years ago

A red photon has a wavelength of 650 nm. An ultraviolet photon has a wavelength of 250 nm. The energy of an ultraviolet photon i

zimovet [89]

The energy of an ultraviolet photon is 2.6 times greater than a red photon.

<h3>Energy of a photon</h3>

The energy of a given photon of light is calculated using the wavelength and the speed of light.

<h3>Energy of a red photon</h3>

The energy of the a red photon at the given wavelength is calculated as follows;

$E_r = \frac{hc}{\lambda} \\\\E_r = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{650 \times 10^{-9}} \\\\E_r = 3.078 \times 10^{-19} \ J$

<h3>Energy of an ultraviolet photon</h3>

$E_r = \frac{hc}{\lambda} \\\\E_r = \frac{6.67 \times 10^{-34} \times 3 \times 10^8}{250 \times 10^{-9}} \\\\E_r = 8.0 \times 10^{-19} \ J$

<h3>The ratio of the two energies</h3>

$= \frac{8.0 \times 10^{-19}}{3.078 \times 10^{-19}} \\\\= 2.6$

Thus, the energy of an ultraviolet photon is 2.6 times greater than a red photon.

Learn more about energy of a photon here: brainly.com/question/7464909

4 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago