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oee [108]
3 years ago
10

The statements below are all true. Some of them represent important reasons why the giant impact hypothesis for the Moon’s forma

tion is taken seriously, and some of them are not relevant to deciding between this and other hypotheses. Sort the statements into the correct bin according to whether or not they provide important support to the giant impact hypothesis.a) models indicate early solar system had mars-size leftover panetesimals.b) moon's composition is similar to earth's mantle.c) moon has a low proportion of easily vaporized ingredients.d) moon lacks active volcanoes.e) moon's distance is gradually increasing.f) moon has synchronous rotation.
Physics
1 answer:
Molodets [167]3 years ago
5 0

Answer:

the order of importance must be     b e a f c

Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

e) True. If the moon is material volatilized from the earth it must train a finite receding speed

a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies

f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

         b e a f c

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How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{9}{40}

f=4.44\ cm

When , R₁= -4, R₂ = 8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{3}{40}

f=-13.33\ cm

When , R₁= 4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{3}{40}

f=13.33\ cm

When , R₁= -4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{9}{40}

f=-4.44\ cm

Hence, The lenses with different focal length are four.

8 0
2 years ago
The boiling point of water at sea level is 100 °c. at higher altitudes, the boiling point of water will be
Scilla [17]
   <span> The boiling point of water at sea level is 100 °C. At higher altitudes, the boiling point of water will be.....
a) higher, because the altitude is greater.
b) lower, because temperatures are lower.
c) the same, because water always boils at 100 °C.
d) higher, because there are fewer water molecules in the air.
==> e) lower, because the atmospheric pressure is lower.
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Water boils at a lower temperature on top of a mountain because there is less air pressure on the molecules.
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3 years ago
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sukhopar [10]
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3 years ago
A change in position of an object relating to time
MaRussiya [10]
The answer is: Motion!

Have a great day
3 0
3 years ago
Read 2 more answers
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

7 0
3 years ago
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