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3 years ago

What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?

Physics

1 answer:

PtichkaEL [24]3 years ago

3 0

Answer:

$\tau=3.3*10^{-6}s$

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

$C*\frac{dV}{dt}+\frac{V}{R}=0$

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

$\frac{dV}{V}=-\frac{1}{RC}dt$

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

$\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt$

Evaluating the integrals:

$ln(\frac{V}{v})=e^{\frac{-t}{RC} }$

natural logarithm to both sides in order to isolate V:

$V(t)=ve^{-\frac{t}{RC} }$

Where the term RC is called time constant and is given by:

$\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s$

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Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

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Work lost to friction:

$W_f =\mu mg(c-x)$

Energy:

$E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

(a) Solve for v:

$v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}$

(b) Solve $\frac{dv}{dx}=0$ for x:

$\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}$

$\frac{dv}{dx}=0$ if:

$\mu g-\frac{k}{m}x = 0$

$x_{max} = \frac{\mu gm}{k}$

$v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}$

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3 years ago

Hooke's Law Practice

Oxana [17]

Answer:sheeExplanation:

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3 years ago

A 3.20 kg block starts at rest and slides a distance d down a frictionless β = 30.0 ◦ incline, where it runs into a spring with

Virty [35]

Answer:

Explanation:

a ) work done by gravitational force

= mg sinθ ( d + .21)

Potential energy stored in compressed spring

= 1/2 k x²

= .5 x 431 x ( .21 )²

= 9.5

According to conservation of energy

mg sinθ ( d + .21) = 9.5

3.2 x 9.8 x sin 30( d + .21 ) = 9.5

d = 40 cm

b )

As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.

mg sin30 = kx

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x = 3.63 cm

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3 years ago

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$

$k=\sqrt{\frac{\frac{MR^2}{2}}{M} }$

$k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }$

$k=\sqrt{{\frac{R^2}{2}}$

$k={\frac{R}{\sqrt{2}}$

$k={\frac{1.20m}{\sqrt{2}}$

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = $\frac{2}{3}MR^2$

$k = \sqrt{\frac{\frac{2}{3}MR^2}{M} }$

$k = \sqrt{\frac{2}{3} R^2}$

$k = \sqrt{\frac{2}{3} }*R$

$k = \sqrt{\frac{2}{3}} *1.20$

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of radius

(I) = $\frac{2}{5}MR^2$

$k = \sqrt{\frac{\frac{2}{5}MR^2}{M} }$

$k = \sqrt{\frac{2}{5} R^2}$

$k = \sqrt{\frac{2}{5} }*R$

$k = \sqrt{\frac{2}{5}} *1.20$

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3 years ago

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vazorg [7]

The speed of sound, c, is given by the Newton-Laplace formula
$c = \sqrt{ \frac{K}{\rho} }$
where
K = bulk modulus
ρ = density

Because the density is constant, the speed of sound is proportional to the square root of the bulk modulus.

Therefore when the bulk modulus increases, the speed of sound increases by the square root of the bulk modulus.

For example, if K is doubled, then
$c = \sqrt{2K} = \sqrt{2} \sqrt{K}$

Answer:
If the bulk modulus increases by a factor of n, then c increases by a factor of √n.

7 0

3 years ago