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fgiga [73]
3 years ago
9

A meterstick is placed on a pivot point of 42.5cm and a 45g mass is hung at the 20cm mark. When released the meterstick remains

in static equilibrium. What is the mass of the meterstick
Physics
2 answers:
Vikki [24]3 years ago
6 0
Hey i dont have an answer but i need the points for finals today. Thank you
nordsb [41]3 years ago
3 0
I don’t know the answer sorry I need points
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Based on the second law of thermodynamics, how would you expect a system to change over time?
deff fn [24]

B. It's randomness would increase

Because the Second Law of Thermodynamics states that as energy is transferred or transformed, more and more of it is wasted. It also states that there is a natural tendency of any isolated system to degenerate into a more disordered state.

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Rocks are elastic too! Explain how elasticity and Harry Reid's elastic rebound theory lead to a better understanding of earthqua
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After the great 1906 San Francisco earthquake, geolophysicistHarry Fielding Reid examined the displacement of the ground surface along the San Andreas Fault. He concluded that the quake must have been the result of the elastic reboundof the strain energy in the rocks on either side of the fault.
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7 0
3 years ago
The distance of Saturn from the sun is:<br><br> &lt; 1 A.U.<br> &gt; 1 A.U.<br> = 1 A.U.
sertanlavr [38]

Answer:

1 astronomical unit is the average distance from the Earth to the Sun; approximately 150 million km. At its closest point, Saturn is 9 AU, and then at its most distant point, it's 10.1 AU. Saturn's average distance from the Sun is 9.6 AU. We have written many articles about Saturn for Universe Today.

Explanation:

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3 years ago
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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
Questions 8 out of 20
MrMuchimi

Answer:

potential energy

Explanation:

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