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zepelin [54]
4 years ago
5

When humans started launching rockets into outer space, they needed to make certain the rockets had enough power to escape Earth

's gravitational field. Which of these would have made it much more difficult for humans to design rockets capable of reaching outer space?
A.
if the Earth was much less massive than it is

B.
if the Moon spun on its axis more quickly

C.
if the Earth was much more massive than it is

D.
if the Moon was much more massive than it is
Physics
1 answer:
GaryK [48]4 years ago
5 0
Tha answer has to be a
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2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<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
  • M is magnification.

{ \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.
4 0
2 years ago
Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white
ArbitrLikvidat [17]

The angles for the first-order diffraction of the shortest and longest wavelengths of visible light are 22.33 ⁰ and 49.46 ⁰ respectively.

<h3>Angle for the first order diffraction</h3>

The angle for the first order diffraction is calculated as follows;

dsinθ = mλ

sinθ = mλ/d

<h3>For shortest wavelength (λ = 380 nm)</h3>

d = 1/10,000 lines/cm

d = 1 x 10⁻⁴ cm x 10⁻² m/cm = 1 x 10⁻⁶ m/lines

sinθ = (1 x 380 x 10⁻⁹)/(1 x 10⁻⁶)

sinθ = 0.38

θ = sin⁻¹(0.38)

θ = 22.33 ⁰

<h3>For longest wavelength (λ = 760 nm)</h3>

sinθ = (1 x 760 x 10⁻⁹)/(1 x 10⁻⁶)

sinθ = 0.76

θ = sin⁻¹(0.76)

θ = 49.46 ⁰

Learn more about diffraction here: brainly.com/question/16749356

#SPJ1

8 0
2 years ago
What is absolute zero​
andreev551 [17]

Answer:

-273.15

Explanation:

3 0
3 years ago
The name dinitrogen tetroxide tells you that this compound contains
Leto [7]
The answer to that is c hope it helps
5 0
4 years ago
How much heat is needed to melt 16.5kg of silver that is initially at 20*c?
zubka84 [21]
The Specific Heat Capacity of silver is 230 J/kgK, melting point is 961.8 C so the difference is 941.8K. Now we simply do q=230J/kgK*16.5kg*941.8K and that is 3 574 131 J
8 0
4 years ago
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