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Burka [1]
3 years ago
5

The definition of agility

Physics
1 answer:
Cloud [144]3 years ago
8 0
If basically means to move quickly
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Convert horsepower to Kilo Newton.
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1 horsepower is equivalent to 0.7457 Kilo Newton. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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In a physics lab a 90 kg student runs up a 1.5 meter tall flight of stairs in 3.0 seconds. The students power rating is approxim
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The power generated by the students will be .
•Work done is equal to
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1587.6 j
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1587.6/3
529.2 j/secs
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How much weight can a 2x2 support horizontally?
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Gennadij [26K]
Upper mantle

The theory of the plate tectonics uses the convection cells in the upper mantle layer as the foundation for the movement of the continents.
To put it simple the process is explained like the convection cells create a lot of pressure, that pressure, supported by the high temperatures and magma, influence the crust above, and manage to crack it on certain places. As the crust cracks, it becomes a separate entity, and under the enormous force from bellow it slowly moves into a particular direction, and that is actually the movement of the tectonic plates.
3 0
2 years ago
Read 2 more answers
In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th
wel

Assuming that the earth is the center of all coordinates. The center of mass of the earth will be located at the coordinate x_1 = 0

To calculate the center of mass of the Earth-moon we require to proceed to the calculations given by,

X_{e-m} = \frac{m_1x_1+m_2x_2}{m_1+m_2}

Where

m_1= Mass of earth

x_1 = Distance to center of mass for earth

m_2 = mass of moon

x_2= Distance from moon to earth (to center of mass)

Replacing the values we have

X_{e-m} = \frac{(6*10^{24})(0)+(7.35*10^{22})(3.8*10^5)}{6*10^{24}+0.0735*10^{24}}

X_{e-m} = \frac{27.93*10^{27}}{6.07*10^{24}}

X_{e-m} = 4.599*10^3km

We extrapolate this procedure to the sun, but now including the three bodies, in this way

X_{e-m-s}=\frac{(m_1+m_2)X_{e-m}+m_3x_3}{m_1+m_2+m_3}

X_{e-m-s}=\frac{(6.07*10^{24})(4.599*10^3)+(2*10^{30})(1.5*10^8)}{6.07*10^{24}+2*10^{30}}

X_{e-m-s}= 1.5*10^8km

So we have the two center of mass:

<em>System Earth-Moon equal to 4.599*10^3km</em>

<em>System Earth-Moon-Sun equal to 1.5*10^8km</em>

7 0
3 years ago
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