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Vanyuwa [196]
2 years ago
11

What scientific process would you be using as you identified the characteristics of the buttery

Physics
1 answer:
jarptica [38.1K]2 years ago
8 0

Answer:

The butterfly and moth develop through a process called metamorphosis.

Explanation:

hoped i helped

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Which of the following statements about colloids is FALSE?
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I believe that the answer to this would be B




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2 years ago
Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th
bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

               the mass of planet B to be, Mₓ = M

               the radius of planet A to be, R₁

               the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁²      - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂²                           [where, R₂ = 2R₁]

   = GM/4R₁²    -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

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6 0
1 year ago
A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water incre
butalik [34]

Answer: a) 2,508

Explanation:

6 0
3 years ago
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kolezko [41]

Based on radiometric dating of Apollo rock samples, the rocks have been detected to be about 4.5 Billion years old.

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8 0
3 years ago
If the velocity and frequency of a wave are both doubled, how does the wavelength change?
Sindrei [870]

The wavelength will remain unchanged.

Explanation:

The velocity v of a wave in terms of its wavelength \lambda and frequency \nu is

v = \lambda\nu (1)

so if we double both the velocity and the frequency, the equation above becomes

2v = \lambda(2\nu) (2)

Solving for the wavelength from Eqn(2), we get

\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}

We would have gotten the same result had we used Eqn(1) instead.

5 0
2 years ago
Read 2 more answers
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