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3 years ago

14

What are two processes that result in rocks being broken down into smaller pieces ?

2 answers:

Leya [2.2K]3 years ago

8 0

Answer: Chemical and physical

Explanation:

meriva3 years ago

5 0

Mechanical and Chemical. (Weathering and erosion)

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calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

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3 years ago

Why do we need body membrane

lukranit [14]

So we have a structured form, but can still move. If we had a cell wall we would be stiff objects since it’s just a cell membrane we can still have movement

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3 years ago

Is the following statement true or false? Because action and reaction forces are equal in magnitude, they will produce the same

Deffense [45]

Answer:

I think it's a true but I'm not the smartest

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4 years ago

How would the period of a simple pendulum be affected if it were located on the moon instead of the earth?

OlgaM077 [116]

Answer:

On moon time period will become 2.45 times of the time period on earth

Explanation:

Time period of simple pendulum is equal to $T=2\pi \sqrt{\frac{l}{g}}$ ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth

As when we go to moon, acceleration due to gravity on moon is $\frac{1}{6}$ times os acceleration due to gravity on earth

So time period of pendulum on moon is equal to

$T_{moon}=2\pi \sqrt{\frac{l}{\frac{g}{6}}}=2\pi \sqrt{\frac{6l}{g}}$ --------eqn 2

Dividing eqn 2 by eqn 1

$\frac{T_{moon}}{T}=\sqrt{\frac{6l}{g}\times \frac{g}{l}}$

$T_{moon}=\sqrt{6}T=2.45T$

So on moon time period will become 2.45 times of the time period on earth

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3 years ago

What are two conditions of equilibrium? write them in word

valentina_108 [34]

In order for a system to be in equilibrium , two conditions must be met. Net force must be 0.

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3 years ago