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S_A_V [24]
3 years ago
6

Facts about Neptune and uranus

Physics
2 answers:
Marat540 [252]3 years ago
8 0

Uranus was originally discovered by Sir William Herschel in 1781.  Neptune is the 8th planet from the sun making it the most distant in the solar system.  Neptune is the smallest of the Ice Giants.  The atmosphere of Neptune is made of Hydrogen, helium, And some methane.  Uranus make one trip around the sun every 84 earth years. Uranus has the coldest temperatures of any planet.  I hope This helps!!!

riadik2000 [5.3K]3 years ago
8 0

Uranus's spin axis is rotated at least 97 or 98 degrees while no other planets has anywhere near Uranus's massive tilt (So crazy!) . Neptune is the farthest planet away from the Sun but surprisingly it is not the coldest planet. Uranus is the coldest planet (probably because of it's axis) but it is not the farthest planet from the sun. (Pluto was out of the race because it wasn't considered as a "real planet" and as a "dwarf planet") All the planets in the solar system are named as Greek gods, Uranus is the " Ancient Greek god of the sky " and Neptune was named after the "Roman god of the sea"(possibly because of it's   bluish color ( or colour ) ) .  


Hope this helps :)

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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

7 0
2 years ago
A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to
SpyIntel [72]
This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

 46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s


3 0
3 years ago
Which best explains why people see carrots as orange
Andru [333]

Beta Carotene is the reason why a carrot is orange. Carotene is a color pigment that is found in large quantities in carrots and it has health benefits as well. The main health benefit of carotene is that our body converts it into vitamin A which is essential.

8 0
3 years ago
Read 2 more answers
One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet
Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

M_{s} = sun’s mass

r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

7 0
2 years ago
Yong bought a can of soda at the pool and left the soda in the Sun while he swam. When Yong came back, the soda can was warm. Wh
Aloiza [94]

Answer:

B. heat energy from the sun

4 0
3 years ago
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