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3 years ago

14

Tara prepared a report to show how the amplitude of waves affects the energy of waves. Is her graphical representation correct?

2 answers:

irga5000 [103]3 years ago

8 0

No, it is not correct because as the amplitude of a wave increases, the energy it carries also increases. This graph shows that as the amplitude increases, the energy decreases.

Pls mark as brain-list

Gelneren [198K]3 years ago

6 0

Answer:

It is not correct because the amplitude of the waves can be bigger than others and the graph can be going up and down

Explanation: I got the question right

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With what speed must you approach a source of sound to observe a 25% change in frequency?

insens350 [35]

Sound source is at rest, you are moving with velocity v, f = frequency, c = speed of sound:

f = f0(1 + v/c)

115 = 100(1 + v/343)
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v = 15*343/100
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3 years ago

Why are there only two elements in the first period of the periodic table?(1 point)

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2 years ago

If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th

Trava [24]

Answer:

- path differnce = 2.18*10^-6

- 1538 lines

Explanation:

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$path\ difference\ = dsin\theta$ (1)

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θ: angle of a diffraction = 0.25°

Then, the path difference is:

$path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m$

- The maximum number of bright lines are calculated by using the following formula:

$m\lambda = dsin\theta$ (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

$m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769$

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8 0

3 years ago

Read 2 more answers

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

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3 years ago