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1 year ago

If its wavelength were doubled, its energy would be If its wavelength were doubled, its energy would be 4E. 2E. 12E. 14E.

Physics

1 answer:

ANEK [815]1 year ago

5 0

The wavelength was doubled, and its energy will be increased by 4 times.

looking at the formula

energy $E = MC^2$

also, $c = \lambda \times \nu$

hence it is clear from above that energy is directly proportional to the square of the wavelength.

hence, The wavelength was doubled, and its energy will be increased by 4 times.

<h3>What is Wavelength?</h3>

The distance over which a periodic wave's shape repeats is known as the wavelength in physics.
It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings.
The spatial frequency is the reciprocal of the wavelength. The Greek letter lambda is frequently used to represent wavelength.
The term wavelength is occasionally used to describe modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

To learn more about wavelength with the given link

brainly.com/question/13533093

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1 year ago

What happens to balloon filled with air when it goes very high attitude from surface of earth why

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The balloon will continue to expand and eventually burst.

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2 years ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is: