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kifflom [539]
4 years ago
8

A layer cake profile of different soil types is exposed when we dig into the earth. Geologists reason that these layers were lai

d down sequentially and that layers lower down have to be older than the layers above them. This is called the law of
CorrectA. superposition
Physics
1 answer:
Len [333]4 years ago
3 0

The layer cake profile of different soil types having older soil type at the bottom of the layer is proved by the law of superposition.

Explanation:

The literal meaning of the word superposition means arranging one over the other. So the first generation soil or rocks will be deposited first on the mantle followed by the coming generation of soils and rocks. Due to different kinds of disasters like tides, earthquake, soil erosion etc., the new soil or rocks coming from any other region will get deposited on the surface of the old soil layer just like a layer cake format.

So the presence of older soil in the bottom most part of the mantle and the youngest soil or rock in the topmost surface of Earth is proved by the law of superposition.

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Two blocks with masses M1 and M2 hang one under the other.
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Answer:

(a)T= M2 × g,    (b)T= (M1 + M2)g,   (c)T= M2 (a + g) and  (d)T=(M1 + M2) (a + g)

Explanation:

M1 is hanged upper and M2 is lower at Rest.

(a) For M2

T2 = Weight of the Body M2= M2 × g

(b) T1 = Weight of the Body M2 + Weight of the Body M2

T1 = M1 g + M2 g = (M1 + M2)g

M1 is hanged upper and M2 is lower at accelerated upwards ( F = T - W)

(c) For M2

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How much work must be done in the car to slow it from 100km/h to 50km/h
Brilliant_brown [7]

Answer:

-96.465m joule

Explanation:

Let m = mass of the car and v1 = initial velocity and v2 = final velocity

Given.

Initial velocity = 100 km/h

final velocity = 50 km/h

What is work done in the car to slow it from 100km/h to 50km/h?

v1=100km/h=27.78m/s

v2=50km/h=13.89m/s

The work done in the car to slow it from v1 to v2.

w=Δk

w=k2-k1

w= \frac{1}{2}m(v2)^{2}- \frac{1}{2}m(v1)^{2}

w=\frac{1}{2} m(v2-v1)^{2}

w=\frac{1}{2}\times m(13.89 -27.78)^{2}

w=\frac{1}{2}\times m(-13.89)^{2}

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w=-96.465m joule.

Therefore, the work done is -96.465m joule

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3 years ago
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