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3 years ago

14

Why are weathering, erosion and deposition a NECESSARY process in the rock cycle?

2 answers:

maria [59]3 years ago

5 0

Answer:

Weathering, erosion, and deposistion are necessary processes in the rock cycle because:

Explanation:

First, start with igneous rocks. magma erupts (Extrusive igneous rocks) or solidifies in the sub-surface of the earth (Intrusive igneous rock). when they are exposed Weathering and erosion occur which is a slow breakdown of rock through the wind, water, or other processes. The weathered pieces (sediments) move to other places by wind or water and get deposited someplace else. When there are enough sediments and there is overburden pressure on these sediments, they become a sedimentary rock. Due to overburden pressure, they become metamorphic rocks. now the thing to understand here is that when metamorphic rocks are exposed, they too undergo weathering and erosion and their pieces also become sedimentary rocks.

Zarrin [17]3 years ago

4 0

Answer:

YESS well it is partly nessary but it depends on the situation

Explanation:

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3 years ago

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at

dybincka [34]

Answer:

The intensity at a spot 71 m away is $4.02*10^{-5} Wm^{-2}$

Explanation:

Given:

Initial intensity, $I_{1}=3.0*10^{-4} Wm^{-2}$ at a distance, $d_{1} = 26 m$

Required:

New intensity, $I_{2} =?$ at a distance, $d_{2} = 71 m$

Using the inverse square law,

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⇒ $I_{1}$ $I_{1}d_{1}^{2} =I_{2}d_{2}^{2}$

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3 years ago

At the end of a race a runner decelerates from a velocity of 11.00 m/s at a rate of 0.300 m/s2. (a) how far does she travel in t

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given that initial velocity is

$v_i = 11 m/s$

deceleration is given as

$a = -0.3 m/s^2$

now we have to find the distance covered in 16 s

$d = v_i * t + \frac{1}{2} at^2$

$d = 11*16 - \frac{1}{2}*0.3*16^2$

$d = 137.6 m$

so it will cover 137.6 m distance

part b)

in order to find the final speed

$v_f = v_i + at$

$v_f = 11 - 0.3*16$

$v_f = 6.2 m/s$

so its speed will be 6.2 m/s

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3 years ago

A bugle can be thought of as an open-end pipe. If a bugle were straightened out, it would be 2.65 m long. a. If the speed of sou

vitfil [10]

Answer

given,

Length of pipe, L = 2.65 m

speed of sound, v = 343 m/s

Pipe is open end Pipe

a) Lowest frequency

condition, for open end pipe

$\lambda_1 = 2 L$

$\lambda_1 = 2\times 2.65$

$\lambda_1= 5.30 m$

For lowest frequency

$f_1 = n\dfrac{v}{\lambda_1}$

n = 1

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f₁ = 64.72 Hz

b) For second frequency in bugle

n = 2

$f_2 =2\times \dfrac{343}{5.30}$

f₂ = 129.43 Hz

for n = 3

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f₃ = 194.16 Hz

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3 years ago