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3 years ago

A wave travels at a constant speed. How does the frequency change if the wavelength increases by a factor of 2?

Physics

2 answers:

Kay [80]3 years ago

8 0

A.The Frequency decreases by a factor of 2.

NARA [144]3 years ago

3 0

Answer : The frequency decreases by a factor of 2.

Explanation :

Given that the wave travels at a constant speed. The speed of the wave is given as :

$v=\nu\times \lambda$

Where

υ is the frequency of the wave

and λ is the wavelength of the wave.

In this case, the speed is constant. So, the relation between the frequency and the wavelength is inverse.

$\nu\propto \dfrac{1}{\lambda}$

If the wavelength increases by a factor of 2, its frequency will decrease by a factor of 2.

Hence, the correct option is (A) " The frequency decreases by a factor of 2 ".

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

3 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the mass of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the lab cart

$v_{2}_{i}$ : is the initial velocity of the object = 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the lab cart

$v_{2}_{f}$ : is the final velocity of the object

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object will be the same, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

brainly.com/question/2141713?referrer=searchResults

brainly.com/question/2400186?referrer=searchResults

I hope it helps you!

4 0

3 years ago

Sound waves produced by a source pass a point five times every second. Which of the following choices correctly describes the pe

Ugo [173]

The period is 1/5 second, and the frequency is 5 Hz.

Explanation:

Let's start by reviewing some definitions about waves:

- The period of a wave is the time taken for the wave to make one complete cycle. It is indicated with T and it is measured in seconds (s)

- The frequency of a wave is the number of complete cycles made by the wave in one second. It is indicated with f and it is measured in Hertz (Hz). It is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where f is the frequency and T is the period.

In this problem, we have a wave that passes a given point five time per second. This means that the number of oscillations per second is five, and so its frequency is:

$f=\frac{5 cycles}{1 sec}=5 Hz$