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Leya [2.2K]
3 years ago
12

Which of the following statements can be concluded (deducted) from the parallel-axis theorem? The area moment of inertia of an a

rea about a noncentroidal axis is always less than that about a centroidal axis The area moment of inertia of an area about a noncentroidal axis is always greater than that about a centroidal axis The area moment of inertia of an area about a noncentroidal axis can be equal to that about a centroidal axis There is no relationship between the area moment of inertia of an area about a noncentroidal axis and that about a centroidal axis O None of the above
Physics
1 answer:
Ratling [72]3 years ago
5 0

Answer:

The statement that we can conclude is that the moment of inertial of any body about the centroidal axis is least of all the moment of inertia's of the body about any arbitrary axis in the body.

Explanation:

According to parallel axis theorem we have

I_{x}=I_{C.G}+Ax^{2}

Where

I_{x} is the moment of inertia about any arbitrary axis.

I_{C.G} is the moment of inertia about centroidal axis.

A is the area of section of which Moment of area is evaluated

'x' = is the distance between the axis and C.G

thus we conclude that I_{C.G} is least of all the moments.

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3 years ago
Read 2 more answers
I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

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Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

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c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

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5 0
3 years ago
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AVprozaik [17]

Answer:

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This is the work done by the friction . So this is heat generated.

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