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3 years ago

What is a mafic rock

1 answer:

5 0

Mafic is an adjective describing a silicate mineral or igneous rock that is rich in magnesium and iron, and is thus a portmanteau of magnesium and ferric. Most mafic minerals are dark in color, and common rock-forming mafic minerals include olivine, pyroxene, amphibole, and biotite

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If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser

Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago

By using fossils and matching layers rock layers can be correlated to eachother.

bija089 [108]

I think that the answer to that is true hope that helps

4 0

3 years ago

Read 2 more answers

You calculate the density of a block of aluminum to be 2.68 g/cm3. You look up the density of a block of aluminum at room temper

Hunter-Best [27]

Answer:

Systematic errors.

Explanation:

The density of the aluminium was calculated by a human and this is not natural but can be due to errors in the calibration of the scale for measuring the weight or taking readings from the measuring cylinder.

Random errors are natural errors. Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment. Systematic errors are due to imprecision or problems with instruments.

3 0

3 years ago

A 235.0 g metal block absorbs 2.44 × 103 J of heat to raise its temperature by 35 K. What is the specific heat of the metal? Sho

andrew-mc [135]

When an object absorbs an amount of energy equal to Q, its temperature raises by $\Delta T$ following the formula
$Q=m C_s \Delta T$
where m is the mass of the object and $C_s$ is the specific heat capacity of the material.

In our problem, we have $Q=2.44 \cdot 10^3 J$ , $m=235.0 g$ and $\Delta T=35 K$ , so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
$C_s = \frac{Q}{m \Delta T}= \frac{2.44 \cdot 10^3 J}{(235.0 g)(35 K)}=0.297 J g^{-1} K^{-1}$

3 0

3 years ago