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3 years ago

Which describes why the magma moves in the direction of the arrows?

Physics

2 answers:

Musya8 [376]3 years ago

7 0

The magma is heated and becomes less dense by the core moving it toward the crust.

Option D.

<u>Explanation:</u>

Magma is a form of molten rock which might be partially molten or might be fully molten. Magma consists of silicate liquid.

This molten part of the rock is either ejected in the form of lava or this molten rock might move from upper layer of Earth to the lower layer of Earth. The crystals which are not molten which get transported into the magma, form bubbles and get separated.

lapo4ka [179]3 years ago

7 0

Answer:

The answer is D. The magma is heated and becomes less dense by the core moving it toward the crust.

Explanation:

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Describing motion verbally with distance and displacement

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Answer: Distance is the actual position traveled by an object. Displacement is the shortest distance covered by the object.

Explanation:

Distance is the actual length covered by an object from one position to another. It is a scalar quantity. It cannot be negative.

Displacement is the shortest distance covered by the object. The displacement can be calculated by subtracting initial distance from the final distance. It is vector quantity. The displacement covered by an object can be zero, negative and positive.

Suppose, the boy travels from home to school then come back to his home. The distance between from his home to his school is 3 km.

In this case, the boy covers the total distance from home to school then to home is 6 km. But the displacement is zero because he comes to same position.

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3 years ago

From Gauss's law, the electric field set up by a uniform line of charge is given by the following expression where is a unit vec

Evgesh-ka [11]

Answer:

$\Delta V=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})$

Explanation:

Using the Gauss Law, we obtain the electric Field for a uniform large line of charge:

$2\pi r L*E=\lambda *L/\epsilon_{o}$

$E=\lambda /\(2 \pi* r *\epsilon_{o})$

We calculate the potential difference from the electric field:

$\Delta V=-\int\limits^{r_{1}}_{r_{2}} E \, dr =-\int\limits^{r_{1}}_{r_{2}} \lambda dr/ (2\pi*r*\epsilon_{o})=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})$

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4 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

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2 years ago

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Answer:

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