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OleMash [197]
3 years ago
8

As streams flow through Stone Mountain, layers of sand build up. Over time, the sand particles form a sedimentary rock called sa

ndstone.
What causes sandstone to change into metamorphic rock at Stone Mountain?

Sandstone is reshaped by erosion from a stream.
Sandstone builds up layers of sand at the mountain.
Sandstone experiences intense heat and pressure.
Sandstone has more sand particles compacted on top of it.
Physics
2 answers:
marta [7]3 years ago
7 0

Answer:

Answer is C

Explanation:

I got it right on edg

Firlakuza [10]3 years ago
3 0

As streams flow through Stone Mountain, layers of sand build up. Over time, the sand particles form a sedimentary rock called sandstone. What causes sandstone to change into metamorphic rock at Stone Mountain? Sandstone experiences intense heat and pressure.

(Correct Answer is above)

Also Mark Brainliest please.

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Feliz [49]

Mechanical waves are oscillation of matter, they are important because they all transfer energy from one place to another. There are 2 types of mechanical waves. A transverse wave where the particles vibrate perpendicular to the direction of energy travel and a longitudinal wave where particle vibrations are parallel to the direction of the energy transfer.


I hope it helps, Regards.

4 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
In an electrostatic field, path 1 between points A and B is twice as long as path 2. The electrostatic work done on a negatively
Elanso [62]

Answer:

W2 = W1

Explanation:

work is independent of the path taken between the points.

8 0
2 years ago
How much energy or stopping power is needed to bring a car to a stop from 100 mph?
Fed [463]
I think 100 mph pushing the car the opposite direction
3 0
3 years ago
A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r =  g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0
3 years ago
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