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3 years ago

PLEASE HELP ME!

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

a) A = 3 cm, b) T = 0.4 s, f = 2.5 Hz,

2) A standing wave the displacement of the wave is canceled and only one oscillation remains

Explanation:

a) in an oscillatory movement the amplitude is the highest value of the signal in this case

A = 3 cm

b) the period of oscillation is the time it takes for the wave to repeat itself in this case

T = 0.4 s

the period is the inverse of the frequency

f = 1 /T

f = 1 /, 0.4

f = 2.5 Hz

2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.

A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.

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Write the meaning of an object has 2 meter length

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Answer:

The object has 2 meter length. This means the length is any quantity with a dimension distance. The definition of the length is how long something is or amount of space. In the given data it is stated that, the object is something that has a length of 2 meter.

<em>Let's take examples to understand. </em>

For example a thread or a table is an object which has a total length of 2 meters.

Another example is something we are measuring it gives us a result of 2 meters of length by using a meter scale or meter tape.

Length is a measure of distance and it is a fundamental quantity. Meter is a international system of units (SI units).

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4 years ago

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

Xelga [282]

Answer:

$\lambda=1282\ nm$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

So,

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

Given, $n_i=5\ and\ n_f=3$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m$

$\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m$

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3 years ago

How to determine the width/thickness of a coin using a screw gauge?

belka [17]

We can calculate the width/thickness of a coin using a screw gauge by placing it in between the teeth of the gauge.

<h3>How we can calculate the thickness using screw gauge?</h3>

We can determine the width/thickness of a coin by using a screw gauge because screw gauge is an instrument that measures thickness of an object that are very thin. First we can placed the coin in between the teeth of the gauge and then move the gauge until the coin can be held tightly. After that note the reading on the scale.

So we can conclude that we can calculate the width/thickness of a coin using a screw gauge by placing it in between the teeth of the gauge.

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2 years ago

Unpolarized light of intensity 0.0288 W/m2 is incident on a single polarizing sheet. What is the rms value of the electric field

galina1969 [7]

Answer:

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Explanation:

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$E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m$

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$E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m$

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3 years ago