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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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Timed! I would really appreciate some help! thank you!

GenaCL600 [577]

Answer:

x = 5[km]

Explanation:

We must convert the time from minutes to hours.

$t=30[min]*\frac{1h}{60min}= 0.5[h]\\$

We know that speed is defined as the relationship between space and time.

$v=x/t$

where:

x = space [m]

t = time = 0.5 [h]

v = velocity [m/s]

Now replacing:

$x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]$

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2 years ago

List things in your house and put them in order of density

ELEN [110]

I don't know what's in your house! lol. try naming things like: ball, lamp, cup, bowl.

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3 years ago

Read 2 more answers

A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g

Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

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3 years ago

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Please help answer both of these !! I’m in a rush :(((

Leno4ka [110]

Answer: search it on browser

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2 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: