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Gennadij [26K]
3 years ago
14

A 616 g block is released from rest at height h0 above a vertical spring with spring constant k = 540 n/m and negligible mass. t

he block sticks to the spring and momentarily stops after compressing the spring 23.3 cm. how much work is done (a) by the block on the spring and (b) by the spring on the block? (c) what is the value of h0? (d) if the block were released from height 2h0 above the spring, what would be the maximum compression of the spring?
Physics
1 answer:
galben [10]3 years ago
5 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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Their linear inertia is equivalent to their masses. Let the inertia of the first moose be m₁ and the second be m₂.

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The ratio of their inertias is 2
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The energy required to completely remove the covalent bond between two
hoa [83]

Answer:

B is the answer. Correct me if I'm wrong

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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

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Answer:

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Explanation:

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Because of the magnets are actually electromagnetics aka what causes them to repel each other the atoms and the electrons will make a force of them pushing away from each other because the two magnetic poles are not north and south
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