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Gnom [1K]
4 years ago
15

What is the moment of inertia i of this assembly about the axis through which it is pivoted?

Physics
1 answer:
vivado [14]4 years ago
7 0

Answer:

I = \frac{1}{12}m_t(2x)^2+m_1x^2+m_2x^2

Explanation:

The moment of inertia for the beam is:

I = \frac{1}{12}m_tL^2

Where m_t is the mass of the beam and L is the lengh of the beam

note:

L = 2x

And for particles I is equal to:

I = MR^2

where M is the mass of the particle and R is the distance between the pivot and the particle.

Finally, the moment of inertia for this assembly is the sum of the moment of inertia of the particles and the beam. So:

I = \frac{1}{12}m_t(2x)^2+m_1x^2+m_2x^2

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The phosphorus cycle is the biogeochemical cycle that describes the movement of phosphorus through the lithosphere,
ludmilkaskok [199]

A

most common source of phosphates in the soil is the weathering of the mineral apatite.

Explanation:

Apatite is a group of phosphate minerals like hydroxylapatite, fluorapatite, and chlorapatite. When apatite rocks weather, they release the phosphate minerals mainly in the form of PO₄ ³⁻ . These minerals become dissolved in water (hydrosphere), where they are readily available to plants and other organisms in the biosphere. The phosphates are taken up and used in biosynthesis. When these organisms die and become buried with sediments, the phosphate gets back to the lithosphere as sedimentary rock.

Learn More:

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5 0
3 years ago
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....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

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....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

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

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0
3 years ago
Which characeristic makes omasis different from diffusion.
Ksju [112]
 <span>haha I used to think biology was so hard, i find it quite easy now. 
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Lina20 [59]
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
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And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
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