If a wave is traveling at 200 m/s and its wavelength is 0.5 meters, what is the frequency of the

Physics

1 answer:

anyanavicka [17]3 years ago

7 0

The frequency of the wave = 400/s

<h3>Further explanation</h3>

Given

v=200 m/s

λ=0.5 m

Required

The frequency of the wave

Solution

Frequency (f): number of waves in one second

The frequency is inversely proportional to the wavelength

$\tt f=\dfrac{v}{\lambda}\\\\f=\dfrac{200~m/s}{0.5~m}\\\\f=400/s$

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Read 2 more answers

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by