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BartSMP [9]
3 years ago
14

What rock type does the principle of superposition apply to and how is the principle of superposition used to study the history

of life on Earth?
Physics
2 answers:
azamat3 years ago
8 0
Your answer would be sedimentary I believe
Strike441 [17]3 years ago
3 0

Answer: Sedimentary rocks

The principle of superposition is the basic principle of geology, which states that on examining the sequence of the strata, the stratum that is younger lies as the top whereas the older one rests at the bottom. This principle applies to the sedimentary rocks. As, the sediments are formed and piled one over the other, the compaction of the sediments occurs due to heat and pressure will result in formation of the sedimentary rocks.

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When will the motion of an object be
mafiozo [28]

Answer:

C. The forces are balanced.

Explanation:

Every answer other than C means the object moved.

5 0
2 years ago
What is the purpose of an man made ecosystem
alexandr402 [8]
Usually to preserve the organisms (maybe the organisms are endangered species or need to be protected in some other way). Another reason could be for the benefit of mankind (for example, an artificial river provides recreational purposes)
3 0
3 years ago
A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi
krok68 [10]

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

                 k = [(m)×(v^2)]/(x^2)

                 k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

                 k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

             v^2 = [(k)(x^2)]/m

                 v =  \sqrt{ [(k)(x^2)]/m}

                 v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

                    = 0.491 m/s

8 0
3 years ago
Read 2 more answers
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of
PIT_PIT [208]

Answer:

B = 4.1*10^-3 T = 4.1mT

Explanation:

In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:

\Phi_B=S\cdot B=SBcos\alpha        (1)

ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2

S: surface area of the circular plate = π.r^2

r: radius of the circular plate = 8.50cm = 0.085m

B: magnitude of the magnetic field = ?

α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°

You solve the equation (1) for B, and replace the values of the other parameters:

B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT

The strength of the magntetic field is 4.1mT

4 0
3 years ago
Part A Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.7 m whose uniform depth is 1.8 m . Express y
Sphinxa [80]

Answer:

17.66 kPa

Explanation:

The volume of water in the swimming pool is the product of its dimensions

V = 30 * 8.7 * 1.8 = 469.8 cubic meters

Let water density \rho = 1000 kg/m^3, and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool

W = mg = \rho V g = 1000 * 469.8 * 9.81 = 4608738 N

The area of the bottom

A = 30 * 8.7 = 261 square meters

Therefore the pressure is its force over unit area

P = F/A = 4608738  / 261 = 17658 N/m^2 or 17.66 kPa

7 0
3 years ago
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