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2 years ago

13

What is fiery about igneous rocks?

2 answers:

xz_007 [3.2K]2 years ago

8 0

It is fiery because it is cooled molten magma

Sveta_85 [38]2 years ago

6 0

Igneous rocks are fiery because they are molten rocks that had solidified

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In which layer of the sun is energy transfferred between atoms?

sasho [114]

Answer:

Core is the layer of the sun is energy transferred between atoms.

Explanation:

4 0

2 years ago

When parts of your body functions as lever systems, what part supplies the force?

mylen [45]

Muscles supply the force. They exert force by getting shorter (contracting).
A muscle can't get longer by itself. There needs to be another muscle
pulling in the opposite direction.

5 0

2 years ago

A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).

yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

$f = n\frac{v}{2L}$

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

$f = n\frac{v}{2L}\\L = \frac{nv}{2f}$

The radius between the two frequencies would be 4 to 5,

$\frac{528Hz}{660Hz}= \frac{4}{5}$

$4:5$

Therefore the frequencies are in the ratio of natural numbers. That is

$4f = 528\\f = \frac{528}{4}\\f = 132Hz$

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

$L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m$

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

$L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m$

Therefore the length of the pipe in the second harmonic is 2.6m

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2 years ago

Roots show ____. geotropism because they grow toward gravitational pull. 1.) positive 2.) neutral 3.) negative 3.)

solniwko [45]

Its the second option if you have Suttons Bay Virtual Online School

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2 years ago

Anti-reflective coatings on lenses use thin-film interference to eliminate the reflection of a particular color. Suppose a glass

Sidana [21]

Answer with Explanation:

We are given that

$n_g=1.55$

$n_f=1.35$

$\lambda=521 nm=521\times 10^{-9} m$

$1 nm=10^{-9} m$

a.We have to find the expression for the minimum thickness the film can have ,t.

Condition for destructive interference

$2nt=(m+\frac{1}{2})\lambda$

$t=\frac{(m+\frac{1}{2})\lambda}{2n}$

For minimum thickness m=0

Then, $t=\frac{\lambda}{4n}$

b.Substitute the values

$t=\frac{521\times 10^{-9}}{4\times 1.35}=96.5nm$

7 0

2 years ago