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2 years ago

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Why do the layers of sedimentary rocks parallel to earth's surface?

1 answer:

Vedmedyk [2.9K]2 years ago

6 0

Answer:

why do the layers of sedimentary rocks parallel to earth's surface?

Why are sedimentary rocks found on or close to Earth's surface?

Explanation: Erosion breaks existing rocks and turns them into sediments. ... When subjected to heat and pressure deeper in the earth sedimentary rocks are changed into igneous and metamorphic rocks. Because erosion only happens on the earth's surface sedimentary rocks are only formed on the earth's surface.

Explanation:

Why do sedimentary rocks have layers and how these layers are formed?

The different groups of sediments could have been deposited through wind, water, ice, and/or gravity at different intervals of time and compacted on top of each other, until they create a sedimentary rock that has several different types of sediments (possibly from different rock types) in the form of layers.

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Two negative charges that are both -3.0 C push each other apart with a force of 19.2 N. How far apart are the two charges ?

fgiga [73]

The Coulomb's force acting between two charges is
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e = 8.99 \cdot 10^9 N m^{-2}C^{-2}$ is the Coulomb's constant, q1 and q2 the two charges, and r the distance between them.

Using $q_1 = q_2 = -3 C$ , we can find the distance between the two charges when the force is F=19.2 N:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99\cdot 10^9 \frac{(-3.0C)^2}{19.2 N} }=6.5 \cdot 10^4 m = 65 km$

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3 years ago

Which best describes the net force on a pair of balanced forces?

Butoxors [25]

The net force of a pair of balanced forces is zero

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3 years ago

Read 2 more answers

Calculate the magnitude of the focal length and power of a spherical mirror that has a radius of 5.00 cm.

yarga [219]

Answer:

2.5 cm

40 D

Explanation:

When the radius of curvature of a lens is divided by 2 we get the focal length of the lens.

Focal length is given by

$f=\dfrac{R}{2}\\\Rightarrow f=\dfrac{5}{2}\\\Rightarrow f=2.5\ cm$

The focal length of the lens is 2.5 cm

When we divide 1 by the focal length in the unit of meters we get the power of a lens

Power of a lens is given by

$P=\dfrac{1}{f}\\\Rightarrow P=\dfrac{1}{0.025}\\\Rightarrow P=40\ D$

The power of the lens is 40 D

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4 years ago

A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The

Goshia [24]

Answer:

$0.629\ \text{rad/s}^2$ counterclockwise

$9.98\ \text{s}$

Explanation:

$r_1$ = Small drive wheel radius = 2.2 cm

$\alpha_1$ = Angular acceleration of the small drive wheel = $8\ \text{rad/s}^2$

$r_2$ = Radius of pottery wheel = 28 cm

$\alpha_2$ = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

$r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2$

The angular acceleration of the pottery wheel is $0.629\ \text{rad/s}^2$ .

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = $60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}$

t = Time taken

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}$

The time it takes the pottery wheel to reach the required speed is $9.98\ \text{s}$

4 0

3 years ago

A force of 20 newtons is exerted on a chair from its right and a force of 12 newtons is applied from its left. What is the net f

jeka94

The net force is the sum of the forces acting on an object ( chair ).
F net = F 1 + F 2
F 1 = 20 N, F 2 = - 12 N
F net = 20 N + ( - 12 N ) = 8 N
The chair will tend to move to the left.

4 0

3 years ago