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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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astra-53 [7]

Option A is the false statement.

Only materials that have never been part of a living thing can be recycled in nature.

In actual all living and non living thing can be recycled in nature.

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3 years ago

The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average r

Dvinal [7]

Answer:

The lifetime of the particle is $\Delta t = 2.6*10^{-25} \ s$

Explanation:

From the question we are told that

The average rest energy is $E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J$

The intrinsic width is $\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J$

The lifetime is mathematically represented as

$\Delta t = \frac{h}{\Delta E}$

Where h is the Planck's constant with a value of $1.055*10^{-34} \ J\cdot s$

substituting values

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$\Delta t = 2.6*10^{-25} \ s$

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2 years ago

Which of the following objects has the greatest density

maks197457 [2]

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2 years ago

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An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2010 × 103 seconds (about

vazorg [7]

Answer:

Radial acceleration of moon is $a_{r} = 2.246\times 10^{-3}$ $\frac{m}{s^{2} }$

Explanation:

Given :

Time period $T = 1.987 \times 10^{6}$ sec

Distance from center of moon to planet $r = 225 \times 10^{6}$ m

From the equation of radial acceleration,

$a_{r} = r\omega ^{2}$

Where $\omega = 2\pi f = \frac{2\pi }{T}$

So $\omega = 3.16 \times 10^{-6} \frac{rad}{s}$

Now moon's radial acceleration,

$a_{r} = 225 \times 10^{6} \times (3.16 \times 10^{-6} )^{2}$

$a_{r} = 2246.76 \times 10^{-6}$

$a_{r} = 2.246\times 10^{-3}$ $\frac{m}{s^{2} }$

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2 years ago

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algol [13]

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2 years ago