All categories

2 years ago

How can you tell if an igneous rock has an intrusive or extrusive origin?

1 answer:

4 0

Intrusive igneous rocks cool down from magma slowly because they form underneath the surface, that will make them have large crystals.

Extrusive igneous rocks cool down from lava rapidly because they form at the surface, so that will make them have small crystals.

You might be interested in

Soccer can be traced back to the early Greeks and Romans.<br> O True<br> O False

Natalija [7]

Can Soccer be traced back to the early Greeks and Romans

Answer:

O False,

Explanation:

Although the game of soccer has been around for more than 2,000 years, soccer as we know it today is traced back to England. The game was once played in ancient China, Greece, Rome, and Japan but with different rules and variations.

8 0

2 years ago

A 7.0-kg rock is subject to a variable force given by the equation f(x)=6.0n−(2.0n/m)x+(6.0n/m2)x2 if the rock initially is at r

stealth61 [152]

For Newton's second law, the force is equal to the product between the mass and the acceleration of the rocket:
$F=ma$
From which we can rewrite the acceleration as
$a(x)= \frac{F}{m} = \frac{1}{m} (6-2x+6x^2)$
where m=7.0 kg.

The velocity of the rocket is the derivative of the acceleration:
$v(x) = \frac{1}{m} (-2+12 x)$
and if we substitute x=9.0 m, we find the rocket velocity after 9.0 m:
$v(9)= \frac{1}{7}(-2+12\cdot 9)=15.1 m/s$

4 0

2 years ago

What is the equivalent resistance of the

BigorU [14]

Answer:

Approximately $111\; {\rm \Omega}$ .

Explanation:

It is given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage $V$ .

Since $R_{1}$ and $R_{2}$ are connected in parallel, the voltage across the two resistors would both be $V$ . Thus, the current going through the two resistors would be $(V / R_{1})$ and $(V / R_{2})$ , respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: $I = (V / R_{1}) + (V / R_{2})$ .

In other words, if the voltage across this circuit is $V$ , the total current in this circuit would be $I = (V / R_{1}) + (V / R_{2})$ . The (equivalent) resistance $R$ of this circuit would be:

$\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}$ .

Given that $R_{1} = 200\; {\Omega}$ and $R_{2} = 250\; {\Omega}$ :

$\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}$ .

7 0

1 year ago

When NASA was communicating with astronauts on the Moon, the time from sending on the Earth to receiving on the moon was 1.33 s.

frez [133]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description, which determine the velocity, such as the displacement of a particle as a function of time, that is to say

$v = \frac{x}{t}\rightarrow x = v*t$