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Dmitry_Shevchenko [17]
3 years ago
9

‼️‼️ Please help, urgent ‼️‼️ (check photo)

Physics
1 answer:
Alex787 [66]3 years ago
5 0

Answer: The force constant k is 10600 kg/s^2

Step by step:

Use the law of energy conservation. When the elevator hits the spring, it has a certain kinetic and a potential energy. When the elevator reaches the point of still stand the kinetic and potential energies have been transformed to work performed by the elevator in the form of friction (brake clamp) and loading the spring.

Let us define the vertical height axis as having two points: h=2m at the point of elevator hitting the spring, and h=0m at the point of stopping.

The total energy at the point h=2m is:

E_{tot}=E_{kin}+E_{pot}\\E_{tot}= \frac{1}{2}mv^2+mg\Delta h = \frac{1}{2}2000 kg 4^2\frac{m^2}{s^2}+2000kg\, 9.8\frac{m}{s^2}2m=55200\,kg\frac{m^2}{s^2}

The total energy at the point h=0m is:

E_{tot}=E_{kin}+E_{pot}+Work=0+0+ Work\\E_{tot} =F_{friction}\Delta h+\frac{1}{2}k (\Delta h)^2=17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2

The two Energy values are to be equal (by law of energy conservation), which allows us to determine the only unknown, namely the force constant k:

17000N\cdot 2m+\frac{1}{2}k\cdot 2^2 m^2 = 55200 \,kg\frac{m^2}{s^2}\\k = \frac{55200-34000}{2}\,\frac{kg}{s^2}=10600\frac{kg}{s^2}

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There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
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Answer:

The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

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The electric potential at a distance d due to 'Q' is:

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Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

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Now,

The Electric field due to charge Q at a distance is given by:

\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

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\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

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