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zheka24 [161]
3 years ago
6

What is the cuplike depression of the hip bone into which the head of the femur fits?

Biology
1 answer:
Kruka [31]3 years ago
8 0

The cup like structure, in which the femur fits into the hip bone is known as Acetabulum.

the acetabulum is referred to a concave shapoed surface of the pelvis, which have a cup like groove. This groove helps in the binding of the femur head to the hip or the pelvic bone.

The round head of the femur gets fits into this cup shaped structure and forms the hip joint. The acetabulum is formed by the contribution of ischium, ilium and the pubis.

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Glaciers are massive bodies of ice that move across the surface of the Earth like rivers. As glaciers move they change the surface of the Earth by wearing away loose rocks and soil and depositing them somewhere else.

Explanation:

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Describe the structure of a phospholipid
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How does a hypothesis help scientists understand the natural world?
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A steel bar has a mass of 4,000 kilograms and a volume of 0.5 cubic meters . Calculate its density. Write your answer as a whole
mixer [17]

The equation for density is:


D=\frac{m}{V}


where the mass is in grams and the volume is in cm^{3}.


We are given 4,000 kilograms, and we know that there are 1,000 grams in 1 kilogram, so we need to multiply to convert to grams:


4,000*1,000=4,000,000g


We are also given 0.5 m^{3}, and we need to convert to cm^{3}. We know that there are 100 cm in 1 m, and since we are talking about cubic volume, we must also raise 100 to the 3rd power: 100^{3}=1,000,000cm. So now we know that there are 1,000,000 cm^{3} in 1 m^{3}. We need to then multiply 1,000,000 cm by 0.5 in order to find the volume that was given in the question:


1,000,000*0.5=500,000cm^{3}


So now that we have the correct units, let's plug them into the equation:


D=\frac{4,000,000g}{500,000cm^{3}}


D=\frac{8g}{cm^{3}}


So now we know that the density is 8g/cubic centimeter.

3 0
3 years ago
This stadium can hold 100,000, or 1x10, people. The
Mekhanik [1.2K]

Answer:

1 x 10^13 stadiums will be needed in this scenario

Explanation:

We are told that

1 stadium holds = 1 × 10^5 people and

Number of iron atoms = 1 × 10^18 atoms

If the stadium carries an equivalent number of atoms as that of people.

We can infer that 1 stadium will carry 1 × 10^5 atoms.

The calculation to determine the number of stadiums would then be 1 × 10^18 divided by 10^5 atoms/stadium which was gotten by dividing the total number of atoms by the number of atoms per stadium.

Number of stadiums = Total number of atoms ÷ Number of atoms per stadium

= 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium

= 1 × 10^13 Stadiums

This means that 1 × 10^18 atoms would occupy 1 × 10^13 stadiums

6 0
3 years ago
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