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3 years ago

About how many known elements are there on Earth? Are they all naturally made?

2 answers:

5 0

There are exactly 92 elements on Earth, all products of nature.

More have been created in laboratory 'atom smashers' and
particle accelerators, but their atoms don't last long ... they fall apart.
(That's why these elements are not found in nature ... the Earth is
more than 4 billion years old, and those complicated atoms of
elements higher than #92 fell apart before now.)

The main parts of the atom are:

-- the "nucleus", in the center, containing neutrons and protons;
the number of protons is what determines what element it's an atom of

-- several electrons, whizzing around the nucleus, like planets in orbits,
but so fast that you can never know exactly where any electron is; they
form kind a fuzzy cloud all around the nucleus;

-- the electrons are far away from the nucleus, so the atom is mostly
empty space;
I've read that if the cloud of whizzing electrons were the size of Texas,
then the nucleus in the center would be the size of a bunch of grapes !

-- most of the time, there are the same number of electrons as the
number of protons in the nucleus, but not always;
it's fairly easy for electrons to get pulled away from an atom atom, or to
temporarily hook up with an atom where it doesn't really belong;
one situation where this happens this happens is when there's an electric
current flowing in the substance ... a lot of electrons are drifting from one
atom to another, generally in the direction of the positive terminal of the battery.
For example: If there's 1 Ampere of current flowing through a piece of wire,
then there are about 6,250,000,000,000,000,000 electrons flowing into
one end of the wire and out the other end, every second.

enot [183]3 years ago

3 0

92 elements, yes, 3-protons, neutrons, & electrons, i dont know the last one

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A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

7 0

3 years ago

An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of

slega [8]

Answer:

$T_{1}=94.9^{o}C$

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

$T_{1}=T_{2}+\frac{q}{kS}$

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

$S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\$

Substitute the given values

$S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m$

The temperature of heater is then:

$T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C$

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

$T_{1}=94.9^{o}C$

5 0

3 years ago

How to calculate displacement?

Blizzard [7]

-- Take a straight ruler.

-- Lay it down with the 'zero' mark at the start point.

-- Rotate it around the start point until the end point is also touching the edge of the ruler.

-- From the marks on the ruler, read the straight-line distance from the start point to the end point.

-- Without moving the ruler, observe and write down the DIRECTION from the start point to the end point.

-- The Displacement is the straight-line distance and direction from the start point to the end point.

4 0

3 years ago

A hungry lion is looking for food. It went 5 miles (mi) before it spotted a zebra to stalk. It took this lion 0.2 hours to get t

siniylev [52]

The lion covered 5 miles in 0.2 hours. With a quick division you can find the speed per hour: 5 / 0.2 = 25 mi/h

4 0

4 years ago

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The launch speed of a projectile is three times the speed it has at its maximum height.what is the elevation angle at launch?

Sphinxa [80]

Answer:

Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.

6 0

3 years ago