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3 years ago

Why does magma rise from depth to the surface of the earth?

Physics

2 answers:

Viefleur [7K]3 years ago

6 0

Answer

Magma is less dense compared to the surrounding rock.

the overlying rock creates pressure which forces the magma to be directed upward.

Explanation:

at high temperatures the magma is liquid form with the high energy which causes the formation of bonds and the pressure build up creates the increase channeling of the liquid.as the temperature decreases the magma moves into the surface

grigory [225]3 years ago

4 0

Explanation:

Any rock will melt once it reaches a high enough temperature. This liquid rock is known as magma when it is beneath the crust of the Earth. If it bubbles to the surface and pushes its way out, it is known as lava. This happens most often in a volcano.

Magma rising from the depth occurs when hot mantle rock rises to shallower depths in the Earth because it is less dense than surrounding rock and because the weight of the overlying rock creates pressure that squeezes magma upward.

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A compass needle acts as a(n)

beks73 [17]

Explanation:

A compass needle acts as a Magnet

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3 years ago

The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the

Studentka2010 [4]

Answer:

$V(t)= 240V* H(t-5)$

Explanation:

The heaviside function is defined as:

$H(t) =1 \quad t\geq 0\\H(t) =0 \quad t$

so we see that the Heaviside function "switches on" when $t=0$ , and remains switched on when $t>0$

If we want our heaviside function to switch on when $t=5$ , we need the argument to the heaviside function to be 0 when $t=5$

Thus we define a function f:

$f(t) = H(t-5)$

The $-5$ term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. ( $H(t-5) =1$ when $t\geq 5$ , so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

$V(t)= 240V* H(t-5)$

I have made a sketch for you, and added it as attachment.

5 0

3 years ago

The sum of kinetic energy of molecules in a body is ______

Margarita [4]

Mechanical energy is the answer

6 0

2 years ago

4. A 1,000-kilogram satellite completes a uniform circular orbit of radius 8.0 x 10 meters as

lesantik [10]

Answer:

Zero

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement of the object

In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so $\theta=90^{\circ}$ and $cos \theta =0$ . Therefore, the work done by gravity is also zero.

5 0

3 years ago

(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t

Alex73 [517]

Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:

Q = m * Lf

Lf = 333,000 J/kg

Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

Q = 4,861.8 J

Now, the entropy change is:

ΔS = 4,861.8 / 273

ΔS = 17.81 J/K

b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

Now the entropy change:

ΔS = 12,430.56 / 373

ΔS = 33.325 J/K

8 0

3 years ago