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3 years ago

15

Metamorphic rock occurs when you take any other type of rock and put them under and ______, and they twist and

meld together.

1 answer:

GrogVix [38]3 years ago

4 0

Answer:

This question appear incomplete

Explanation:

This question appear incomplete because of the absence of options. However, metamorphic rocks are rocks that are formed from other pre-existing rocks under heat and pressure (both are usually high), causing them to twist and melt together. Metamorphism simply means a change in form of something and that's what happens here also.

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There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee

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The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

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Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0

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Answer:

20 hertz of frequency produced.

Explanation:

$\boxed{frequency = \frac{1}{period} }$

Here we will find frequency and period should be in second, here given: 0.05 seconds

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$\boxed{frequency = \frac{1}{0.05} }$

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2 years ago

A stunt woman of mass m falls into a net during the filming of an action movie. Assume she experiences upward acceleration magni

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Answer:

M= F^n / a+g

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3 years ago

Read 2 more answers

Water (density = 1 ´ 103 kg/m3) flows at 10 m/s through a pipe with radius 0.030 m. the pipe goes up to the second floor of the

Hatshy [7]

density of water = $1000 kg/m^3$

velocity of flow = $10 m/s$

radius of pipe = $0.030 m$

Height of second floor = $2 m$

Now we can use here Bernuoli's Equation to find the speed of water flow at second floor

$P_1 + 1/2\rho v_1^2 + \rho g h_1= P_2 + 1/2 \rho v_2^2 + \rho g h_2$

$P + 1/2 * 1000 * 10^2 + 1000* 9.8 * 0 = P + 1/2 * 1000 * v^2 + 1000*9.8*2$

$v = 7.8 m/s$

Now in order to find the radius of pipe we can use equation of continuity

$A_1 v_1 = A_2 v_2$

$\pi *0.030^2 * 10 = \pi * r^2 * 7.8$

$r = 0.034 m$

So radius of pipe at second floor is 0.034 meter

3 0

3 years ago

An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the

OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

$m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2$

Through the aforementioned formula we will have to

$v_f^2-v_i^2 = 2ax$

The particulate part of the rest, so the final speed would be

$v_f^2 = 2gx$

$v_f=\sqrt{2(9.8)(3.1)}$

$v_f = 7.79m/s$

Now from Newton's second law we know that

$F = ma$

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

$F = m\frac{dv}{dt}$

Replacing we have that,

$F = (89)\frac{7.79}{0.5}$

$F = 1386.62N$

Therefore the force that the water exert on the man is 1386.62

3 0

3 years ago