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3 years ago

The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency?

Physics

2 answers:

maria [59]3 years ago

8 0

F=v/wavelength f=1.2/60=50Hz.

Radda [10]3 years ago

3 0

Answer : Frequency = 50 Hz

Explanation :

It is given that,

Wavelength of a wave on a string, $\lambda=1.2\ m$

The speed of the wave is, v = 60 m/s

We have to find the frequency of the wave.

We know that the speed of the wave is given by the product of the frequency of the wave and the wavelength.

$v=\nu \lambda$

$\nu=\dfrac{v}{\lambda}$

$\nu=\dfrac{60\ m/s}{1.2\ m}$

$\nu=50\ Hz$

So, the frequency of the wave is 50 Hz.

Hence, this is the required solution.

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A long solenoid has 103 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.60 cm perp

Aneli [31]

Answer:

0.23348 A

Explanation:

B = Magnetic field

v = Velocity of electron = $1.38\times 10^7\ m/s$

q = Charge of electron = $1.6\times 10^{-19}\ C$

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius of circle = 0.026 m

N = Number of turns = 103 turns/cm = $103\times 100\ turns/m$

I = Current

The magnetic and centripetal force will be balanced

$Bqv=m\frac{v^2}{r}\\\Rightarrow B=\frac{mv}{qr}$

The magnetic field in solenoid is given by

$B=N\mu_0 I\\\Rightarrow I=\frac{B}{N\mu_0}$

From the first equation

$I=\frac{\frac{mv}{qr}}{N\mu_0}\\\Rightarrow I=\frac{mv}{N\mu_0qr}\\\Rightarrow I=\frac{9.11\times 10^{-31}\times 1.38\times 10^7}{103\times 100\times 4\pi \times 10^{-7}\times 1.6\times 10^{-19}\times 0.026}\\\Rightarrow I=0.23348\ A$

The current in the solenoid is 0.23348 A

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