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irina1246 [14]
2 years ago
10

On which of saturn’s moons did the cassini-huygens probe land in 2004, providing our first view of the varied and active surface

?
Physics
1 answer:
yawa3891 [41]2 years ago
4 0

On Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

To find the answer, we have to know more about the Cassini-Huygens Mission.

<h3>What is Cassini-Huygens mission?</h3>
  • Before arriving at its final destination of Saturn in 2004 and beginning a series of flybys of Saturn's moons, the spacecraft contributed to studies of Jupiter for six months in 2000.
  • In the same year, it launched the Huygens probe to explore Titan's atmosphere and surface makeup on Saturn's moon.
  • During its second extended mission, Cassini sailed between the rings, entered the planet's atmosphere, and obtained the first measurements of a whole seasonal period for Saturn and its moons.

Thus, we can conclude that, on Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

Learn more about the Cassini-Huygens mission here:

brainly.com/question/27907891

#SPJ4

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2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
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<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
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