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erastovalidia [21]
3 years ago
14

Please help me with this question someone

Physics
2 answers:
Gekata [30.6K]3 years ago
8 0
From the answers provided, I believe the possible answer would be the last option, silicon, oxygen, and one or more metals. Here's my reasoning: the most abundant mineral group found in the Earth's crust is the silicate group. The silicate materials contain both oxygen and silicon. Silicates are the most common minerals in the rock-formation process, and it has, in fact, been estimated that they make up 75 to 90 percent of the Earth's crust. From this piece of evidence, I can guess that the answer will possibly be D, silicon, oxygen, and one or more metals.
It should also be noted that the additional elements that combine with the silicon-oxygen tetrahedron are involved with the other elements commonly found in the Earth's crust and mantle. They are aluminum, calcium, iron, magnesium, potassium and sodium.
Marianna [84]3 years ago
5 0
The last option, silicon, oxygen, and one or more metals
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Question 23
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If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

с  Stay the same

Explanation:

  • The Mass of an object or body does not affect the acceleration due to gravity in any kind of way.
  • Light weight objects accelerate more slowly than the heavy objects because when the forces other than the gravity also plays a major role.
  • Mass increases of a body when an object has higher velocity or the speed.
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3 0
3 years ago
If a population of 1,000 individuals has 160 rr genotypes, what is the frequency of the rr genotype?
lesya692 [45]
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= 0.16 or 16%
7 0
3 years ago
Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

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so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

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3 years ago
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Answer:

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Explanation:

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5 0
3 years ago
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