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Alexxx [7]
4 years ago
5

2. A mass M is suspended from a frictionless hinge on a massless stick of length b. A massless rod is held perpendicular to the

first rod (making ski-lift ``T-bar'') and another mass m slides frictionlessly along the second rod. Both rods pivot in the plane of the T-bar. Write equations of motion for both masses under the hilarious assumption that the masses can pass through each other without interacting in any way. What are the equations for small oscillations of m and M
Physics
1 answer:
stealth61 [152]4 years ago
6 0

Answer: f=Mg for mass M

And f=mg for m

Equation for small oscillation of m is T=2Π√m/k

For M it will be

T=2Π√M/k

Explanation: the force acting on mass M will be f=Mg in a downward direction f=mg in a downward direction.

Net force acting downward will be f=g(M+m).

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
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Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

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