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3 years ago

15

The flash unit in a camera uses a special circuit to "step up" the 3.0 V from the batteries to 360 V , which charges a capacitor

. The capacitor is then discharged through a flashlamp. The discharge takes 12 μs , and the average power dissipated in the flashlamp is 1.0×105 W .What is the capacitance of the capacitor? use 2 sigfigsI have tried 0.27 F and 0.26 F but so far both are wrong on mastering physics.

1 answer:

aksik [14]3 years ago

7 0

Answer:

The capacitance of the capacitor is $1.85*10^{-5}F$

Explanation:

To solve this exercise it is necessary to apply the concepts related to Power and energy stored in a capacitor.

By definition we know that power is represented as

$P = \frac{E}{t}$

Where,

E= Energy

t = time

Solving to find the Energy we have,

$E = P*t$

Our values are:

$P = 1*10^5W$

$t = 12*10^{-6}s$

Then,

$E= (1*10^5)(12*10^{-6})$

$E = 1.2J$

With the energy found we can know calculate the Capacitance in a capacitor through the energy for capacitor equation, that is

$E=\frac{1}{2}CV^2$

Solving for C=

$C = \frac{1}{2}\frac{E}{V^2}$

$C = 2\frac{1.2}{(360^2-3^2)}$

$C = 1.85*10^{-5}F$

Therefore the capacitance of the capacitor is $1.85*10^{-5}F$

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Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

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Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

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Answer:

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Explanation:

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