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JulsSmile [24]
3 years ago
6

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

Physics
1 answer:
Murrr4er [49]3 years ago
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.
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Friction occurs when microscopic hills and valleys stick together<br><br> true or false
Reika [66]

Answer:

true

Explanation:

Friction occurs because no surface is perfectly smooth. Rougher surfaces have more friction between them. Heavier objects also have more friction because they press together with greater force. Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.

4 0
3 years ago
Read 2 more answers
A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave
Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction \mu =0.4

Consider a small element at an angle \theta having an angle of d\theta

Normal Force=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}

N=T\cdot d\theta

Friction f=\mu \times Normal\ Reaction

f=\mu \cdot N

and T+dT-T=f=\mu Td\theta

dT=\mu Td\theta

\frac{dT}{T}=\mu d\theta

\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta

\frac{T_2}{T_1}=e^{\mu \pi}

\frac{T_2}{T_1}=e^{0.4\times \pi }

\frac{T_2}{T_1}==e^{1.256}

\frac{T_2}{T_1}=3.51

7 0
3 years ago
What does an electric circuit mean?​
nikitadnepr [17]

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Explanation: Hope this helped! <3

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3 years ago
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patriot [66]

Answer:

answer is friction. MCQ A is answer

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3 years ago
A 0.095 kg tennis ball is traveling to the right at 40 m/s when it bounces of a wall and travels in the opposite direction it ca
Mashutka [201]

given that

mass of ball = 0.095 kg

initial velocity of ball towards the wall = 40 m/s

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now change in momentum is given as

\Delta P = m(v_f - v_i)

\Delta P = 0.095(30 - (-40))

\Delta P = 6.65 kg m/s

So change in momentum will be 6.65 kg m/s

3 0
3 years ago
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