Sandstone is a sedimentary rock. Which two processes cause sand particles to form sandstone?

Physics

2 answers:

slava [35]2 years ago

8 0

Answer:

number 2 is correct

Explanation:

cementation as icons are deposited by fluids to form a compound that hardens looses sedimentary rocks .

compaction is the density of sedimentary rocks on edge of them, are forced together though sediments mark me brainest add and follow me

yuradex [85]2 years ago

6 0

Answer:

Compaction and cementation

Explanation:

Cementation: As ions are deposited by fluids to form a compound that hardens loose sedimentary rocks.

Compaction: As the density of sedimentary rocks on edge of them are forced together through sediments.

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Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of

WINSTONCH [101]

a₀). You know ...
       -- the object is dropped from 5 meters
   above the pavement;
         -- it falls for 0.83 second.

a₁). Without being told, you assume ...
   -- there is no air anyplace where the marshmallow travels,
   so it free-falls, with no air resistance;
   -- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .

b). You need to find how much LESS than 5 meters
   the marshmallow falls in 0.83 second.

c). You can use whatever equations you like.
   I'm going to use the equation for the distance an object falls in
   ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d). To see how this all goes together for the solution, keep reading:

The distance that an object falls in ' T ' seconds
when it's dropped from rest is

   (1/2 G) x (T²) .

On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall

   (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

   (5.00 - 3.38) = 1.62 meters

above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

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2 years ago

A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw

insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

$[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a$ which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

$[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a$

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

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2 years ago

A person travelled in the -x direction at a constant speed of 2.5 m/s. The person's final position is -10.0 m, and the initial p

Nostrana [21]

Answer:

Explanation:

We have given the final position = -10 m on x axis

And the initial position =10 m

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