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3 years ago

Which set of characteristics best describes igneous rock?

1 answer:

7 0

Some of the main minerals in igneous rocks are feldspar, quartz, olivine and mica. The size of the minerals depends on the depth of the magma that formed the rock. Deeper magma cools more slowly and forms larger crystals. Rocks that cool over a few months have microscopic mineral grains and are called extrusive. Rocks that cool over thousands of years have small to medium grains and are called intrusive. Rocks that cool over millions of years have large pebble sized grains and are called plutonic.

Granite and basalt make up the majority of igneous rocks. Basalt is dark and fine-grained with minerals rich in magnesium and iron. It is either extrusive or intrusive and is the primary rock on the ocean floor. Granite is light and coarse-grained and rich in feldspar and quartz. It is plutonic and less dense than basalt. Granite is found nearly everywhere beneath the continents.

The word "igneous" comes from the Latin word "fire" and is related to the melting process that forms these rocks.

Hope I helped! :)

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Beryllium has a nucleus composed of protons and neutrons. Given the data, how many neutrons are in a typical Beryllium nucleus?

son4ous [18]

16 because you multipy it first

5 0

3 years ago

Match each prefix to the correct multiple of 10. 1. centi- 10 2. kilo- 100 3. deka- 1000 4. milli- 1/100 5. hecto- 1/10 6. deci-

Mice21 [21]

Explanation :

There are some basic metric conversions.

Prefix Multiple of 10

centi $\frac{1}{100}$

Kilo 1000

deka 10

milli $\frac{1}{1000}$

hecto 100

deci $\frac{1}{10}$

For example :

$1cm=\frac{1}{100}m\\\\1Km=1000m\\\\1mm=\frac{1}{1000}m\\\\1Km=10\text{ hecto meter}\\\\1Km=100\text{ deka meter}$

7 0

3 years ago

Read 2 more answers

Question 4 What would be the UCS for Tab the dog? O The shock O The flowers and fence O Fear O The boundary Question 5

Harrizon [31]

The unconditioned stimulus (UCS) for Tab the dog is: B. The flowers and fence.

<h3>The types of stimuli.</h3>

In Science, the two (2) stimuli that are repeatedly paired in classical conditioning include:

Conditioned stimulus

Unconditioned stimulus

<h3>What is an unconditioned stimulus?</h3>

An unconditioned stimulus can be defined as a stimulus that is capable of naturally triggering a response, before or without any conditioning while it elicit a specific response without learning.

This ultimately implies that, an unconditioned stimulus leads to an automatic response in a living organism such as a dog.

In this scenario, the unconditioned stimulus (UCS) for Tab the dog is simply the flowers and fence.

Read more on unconditioned stimulus here: brainly.com/question/24868138

3 0

2 years ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

$V'=\frac{Q}{C'}$

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V$

So, the new voltage is

$V'=2 (12.0 V)=24.0 V$

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

$A=\pi r^2$

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

$A'=\pi (2r)^2 = 4 \pi r^2 = 4A$

While the separation between the plate was unchanged (d); so, the new capacitance will be

$C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C$

So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

$V'=\frac{12.0 V}{4}=3.0 V$

3 0

3 years ago

A student uses a spring with a spring constant of 130 N/m in his projectile apparatus. When 56 J of

Triss [41]

Explanation:

$potentional \: energy \: = \frac{1}{2} k {x}^{2} \\ 56 = \frac{1}{2} \times 130 \times {x}^{2} \\ {x}^{2} = \frac{56}{65} \\ x = \sqrt{ \frac{56}{65} } meter$

plz..

mark it as a brilliant answer

4 0

3 years ago