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Marysya12 [62]
3 years ago
12

WILL MARK BRAINLIEST Why do we see sedimentary rocks more often than igneous and metamorphic rocks?

Physics
1 answer:
Arisa [49]3 years ago
8 0

Answer:

because they are the rocks that line the surface of our planet ​

Explanation:

We see sedimentary rocks more than other rock types because they are the rocks that line the surface of our planet.

Sedimentary rocks typically form the earth cover due to the way they are formed.

  • These rocks are produced by the weathering, transportation and deposition of sediments within a basin.
  • In this basin, the sediment is lithified and converted to sedimentary rocks.
  • These processes are driven by the external heat engine
  • Therefore, it is confined to the surface.
  • Igneous and metamorphic rock's processes are confined to the subsurface.
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What conditions can create the largest waves in the ocean. this is for science. i need done asap.
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The answer is strong winds, i hoped this helped.

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7 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
Aloiza [94]
In other words a infinitesimal segment dV caries the charge 
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<span>drop higher order terms </span>
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<span>To get total charge integrate over the whole volume of your object, i.e. </span>
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<span>With given parameters: </span>
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<span>= 3.77×10⁻⁸C </span>
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