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3 years ago

7

A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 3.6 m, t

his rod will have the same period as a simple pendulum of length ____ cm. Round your answer to the nearest whole number.

1 answer:

salantis [7]3 years ago

3 0

Answer:

The length of the simple pendulum is 2.4 meters.

Explanation:

Time period of simple pendulum is given by :

$T=2\pi\sqrt{\dfrac{L}{g}}$

L is the length of pendulum

The time period of the rope is given by :

$T=2\pi\sqrt{\dfrac{2L'}{3g}}$

L' is the length of the rod, L' = 3.6 m

It is given that, the rod have the same period as a simple pendulum and we need to find the length of simple pendulum i.e.

$2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{2L'}{3g}}$

On solving the above equation as :

$\dfrac{L}{g}=\dfrac{2L'}{3g}$

L = 2.4 m

So, the length of the thin rod that is hung vertically from one end and set into small amplitude oscillation 2.4 meters. Hence, this is the required solution.

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A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

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Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

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where

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2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

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t is the time

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Therefore, at time $t=1 \mu s$ , we have:

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3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

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$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

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