The three types of stress that act on Earth's rocks are compression, tension, and

1 answer:

3 0

The three types of stress that act on Earth's rocks are compression, tension and shear. Among all the options that are given in the question, the correct option is option "D". These kind of stress action creates the rocks to break in a natural way. The earth's rocks sometimes collide with one another, pull apart from each other and sometimes slide against each other. Whenever two rocks collide with each other, they create a compression force. When the two rocks of the earth tries to pull away from each other, it creates a tension force. The sliding of two earth's rocks creates a shearing force.

You might be interested in

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4

DENIUS [597]

Answer:

Is the equation for Ec=1/2 m(Dv)^2 where Dv is the difference between the angular speed & the areolar speed?

5 0

3 years ago

A 1200-kilogram automobile in motion strikes a 0.0001-kilogram insect. As a result, the insect is accelerated at a rate of 100 m

nadya68 [22]

Answer:

force = 1 × $10^{-2}$ N

Explanation:

given data

automobile mass = 1200 kg

insect mass = 0.0001 kg

insect accelerated = 100 m/s²

to find out

magnitude of the force the insect exerts on the car

solution

we get here force the insect exerts that is express as

force = mass × acceleration ............1

put here value we get

force = 0.0001 × 100 m/s²

force = 1 × $10^{-2}$ N

6 0

2 years ago

Read 2 more answers

The Red Sea is widening at a rate of 1.25 centimeters per year. How many years will it take to widen another 812.5 centimeters?

Pani-rosa [81]

E 812 Jak’s now fmfdkc

5 0

2 years ago

I dolt understand what to do next please help

lidiya [134]

Divide 24 by 12.
24/t = 12
24/12 = t
24/12 = 2
t = 2

6 0

2 years ago

) A 1000 kg car travelling at 50 m/s slams into a 1500 kg parked truck in an inelastic manner as

solmaris [256]

(a)

KE = m v^2 / 2 = (1200 kg)(20 m/s)^2 / 2 = 240,000 J

(b)

The energy is entirely dissipated by the force of friction in the brake system.

(c)

W = delta KE = KEf - KEi = (0 - 240,000) J = -240,000 J

(d)

Fd = delta KE

F = (delta KE) / d = (-240,000 J) / (50 m) = -4800 N

The magnitude of the friction force is 4800 N.

8 0

3 years ago