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konstantin123 [22]
3 years ago
7

What types of elements are useful for dating materials?

Physics
2 answers:
Alekssandra [29.7K]3 years ago
4 0

the nswer is for sure radioactive

Semenov [28]3 years ago
3 0
Well im not sure if this is the correct dating materials but here are some examples of Fundamentals of radiometric dating<span>Radioactive decay.
Accuracy of radiometric dating.
Closure temperature.
The age equation.
Uranium–lead dating method.
Samarium–neodymium dating method.
Potassium–argon dating method.
<span>Rubidium–strontium dating method.</span></span>
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The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t
Vikki [24]

Answer:

  • The energy that must be added to the electron to move it to the third excited state is  -1.153 eV
  • The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV

The energy that must be added to the electron to move it to the third excited state is  -1.153 eV

For fourth excited state, n = 5

E₅ = E₁ /5²

E₄ = (1.23 eV) / 25

E₄ = 0.049 eV

Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV

The energy that must be added to the electron to move it to the fourth excited state is  -1.181 eV

5 0
3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
Photosynthesis transforms molecules of water and carbon dioxide into Molecules of?
brilliants [131]
Into molecules of sugar and oxygen.
The complete reaction of the photosynthesis is in fact:
6CO_2+6H_2O \rightarrow C_6H_{12}O_6+6O_2
and the energy of the light coming from the sun is also used to make the reaction possible.
7 0
3 years ago
A toy car has a 1.5 A current, and its resistance is 2 Ω. How much voltage does the car require?
Brums [2.3K]

Answer:

Battery Voltage

Explanation:

Voltage refers to the amount of electrical potential your battery holds. The standard automotive battery in today's vehicles is a 12-volt battery. Each battery has six cells, each with 2.1 volts at full charge. A car battery is considered fully charged at 12.6 volts or higher. hope this helps you :)

4 0
3 years ago
15. Missy Diwater, the former platform diver or the Ringling Brother's Circus had a speed of 25m/s when she hit the
timurjin [86]
K.E.= 1/2 x MV^2 = 1/2 x 40(kg) x (25x25) =12500J
5 0
2 years ago
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