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4 years ago

What is the difference between regional metamorphism and contact metamorphism?

1 answer:

4 0

Contact metamorphism is a type of metamorphism where rock minerals and texture are changed, mainly by heat, due to contact with magma. Regional metamorphism is a type of metamorphism where rock minerals and texture are changed by heat and pressure over a wide area or region.

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A tank has a gate that automatically opens if the water levelhis high enough. The gate has a squarecross section of side1m and c

umka2103 [35]

Answer:

The gate will open if the height of water is equal to or more than 0.337m.

Explanation:

From the diagram attached, (as seen from the reference question found on google)

The forces are given as

Force on OA

$F_1=P A_1\\F_1=\rho g \bar{h} A_{OA}$

Here

ρ is the density of water.
g is the gravitational acceleration constant
$\bar{h}$ is the equivalent height given as

$\bar{h}=h+\frac{0.6}{2}\\\bar{h}=h+0.3$

$A_{OA}$ is the area of the OA part of the door which is calculated as follows:

$A_{OA}=L\times W\\A_{OA}=1\times 0.6\\A_{OA}=0.6 m^2$

The Force is given as

$F_1=0.6\rho g[h+0.3]$

Force on OB

$F_2=P A_2\\F_2=\rho g \bar{h} A_{OB}$

Here

ρ is the density of water.
g is the gravitational acceleration constant
$\bar{h}$ is the equivalent height given as

$\bar{h}=h+0.6+\frac{0.4}{2}\\\bar{h}=h+0.8$

$A_{OA}$ is the area of the OB part of the door which is calculated as follows:

$A_{OB}=L\times W\\A_{OB}=1\times 0.4\\A_{OB}=0.4 m^2$

The Force is given as

$F_2=0.4\rho g[h+0.8]$

Now the moment arms are given as

$\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+\frac{\frac{1}{12}\times 0.6^3 \times 1}{0.6 \times[h+0.3]}\\\bar{y}_a=h+0.3+\frac{0.03}{h+0.3}$

$\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+\frac{\frac{1}{12}\times 0.4^3 \times 1}{0.4 \times[h+0.8]}\\\bar{y}_b=h+0.8+\frac{0.0133}{h+0.8}$

Taking moment about the point O as zero

$F_1(h+0.6-\bar{y}_a)=F_2(\bar{y}_b-h+0.6)\\F_1(h+0.6-h-0.3-\frac{0.03}{h+0.3})=F_2(h+0.8+\frac{0.0133}{h+0.8}-h-0.6)\\F_1(0.3-\frac{0.03}{h+0.3})=F_2(0.2+\frac{0.0133}{h+0.8})\\0.6\rho g[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4\rho g[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.6[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m$

So the gate will open if the height of water is equal to or more than 0.337m.

3 0

3 years ago

Select the correct answer.

djverab [1.8K]

I believe it’s estrogen

3 0

4 years ago

Read 2 more answers

You place a box weighing 289.9 N on an inclined plane that makes a 35.9 degree angle with the horizontal. Compute the component

Ksivusya [100]

Answer:

Fgparallel = 170N

Explanation:

Acting down the incline would be the paralell force. The Force of gravity on an incline for the parallel portion is mgsinθ.

Fg parallel = mgsinθ

mg is 289.9, as that is the weight. θ is 35.9

Fgpar = 289.9sin35.9

Fgparallel = 170N

5 0

3 years ago

How do the gravitational potential energy and the kinetic energy of the ball change as the ball rolls down the ramp?

belka [17]

Explanation:

The gravitational potential energy is given by :

P = mgh

The kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2$

As the ball reaches the bottom of the ramp, its potential energy decreases and kinetic energy increases.

It imply that, when the ball at the top most height, its gravitational potential energy is maximum and zero kinetic energy and when ball reaches the bottom of the ramp, it will have maximum kinetic energy and zero potential energy.

3 0

3 years ago

7. A car stops at a red light. The light turns green

Readme [11.4K]

Answer:

Graph C

Explanation:

This is the answer because it is the only one that shows the vehicle accelerate to a constant speed before stopping and slowing down.

8 0

2 years ago