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3 years ago

1. The process of producing energy by utilizing heat trapped inside the earth's

Physics

2 answers:

babymother [125]3 years ago

6 0

Answer:

Explanation:

The process is called Geo-Thermal energy because it's an inexhaustible source of energy

earnstyle [38]3 years ago

3 0

Explanation: The process of producing energy by utilizing heat trapped inside the earth surface is called Geo thermal energy. ... Explanation: The heat is apparent from the increase in temperature of the earth with increasing depth below the surface.

Answer ) Geo thermal energy.

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The diameter of a copper atom is approximately 2.28e-10 m. The massof one mole of copper is 64 grams. Assume that the atoms area

KIM [24]

1) Mass of one copper atom: $1.063\cdot 10^{-22} kg$

2) There are $9.33\cdot 10^{24}$ atoms in the cube

3) Mass of the cubical block: 992 kg

Explanation:

We are told here that the mass of one mole of copper is

$M=64 g$

for

$n=1 mol$ (number of moles)

We also know that the number of atoms inside 1 mole of substance is equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

This means that $N_A$ atoms of copper have a mass of M = 64 g. Therefore, we can find the mass of one copper atom by dividing the total mass by the avogadro number:

$m=\frac{M}{N_A}=\frac{64}{6.022\cdot 10^{23}}=1.063\cdot 10^{-22} kg$

We are told that the diameter of a copper atom is

$d=2.28\cdot 10^{-10} m$

We can assume that the atoms are arranged in a cube, and that they are all attached to each other; so the side of the cube can be written as size of one atom multiplied by the number of atom per side:

$L=Nd$

where

N is the number of atoms (rows) in one side of the cube

Since the side of the cube is

L = 4.8 cm = 0.048 m

We find N:

$N=\frac{L}{d}=\frac{0.048}{2.28\cdot 10^{-10}}=2.11\cdot 10^8$

This is the number of atom rows per side; therefore, the total number of atoms in the cube is

$N^3=(2.11\cdot 10^8)^3=9.33\cdot 10^{24}$

The total mass of the cubical block of copper will be given by the mass of one atom of copper multiplied by the total number of atoms, so:

$M= N^3 m$

where:

$N^3 = 9.33\cdot 10^{24}$ is the number of atoms in the cube

$m=1.063\cdot 10^{-22} kg$ is the mass of one atom

Therefore, substituting, we find:

$M=(9.33\cdot 10^{24})(1.063\cdot 10^{-22})=992 kg$

So, the mass of the cubical block is 992 kg.

Learn more about mass and density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

3 0

3 years ago

What might be collected to monitor the amount of excess nitrogen in the local water system

Debora [2.8K]

Answer:

If excess nitrogen is found in the crop fields, the drainage water can introduce it into ... have aquifers that can supply a lot of freshwater very near the land surface.

4 0

3 years ago

A particle is moving along the x-axis so that its position at any time t is greater than and equal to 0 is given by x(t)=2te^-t?

erastovalidia [21]

For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left

6 0

3 years ago

What does physical mean?

kirill [66]

Answer:

relating to the body as opposed to the mind.

8 0

3 years ago

Read 2 more answers

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are: