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Marina86 [1]
3 years ago
10

The table lists several radioisotopes that have technological uses. Which radioisotope would be used by both archaeologists and

botanists (scientists who study plants)? uranium-238 carbon-14 iodine-131 technetium-99
Physics
2 answers:
ElenaW [278]3 years ago
8 0

Answer:

B. carbon-14

Explanation:

Vinil7 [7]3 years ago
5 0
Carbon-14
Since carbon can be found in plants and items that are studied by archaeologist 
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What if is a forecasting game, below are actions on your document and all you have to do is predict what will happen after the a
Talja [164]

1. Triple click on the paragraph.

By extension, the triple click can allow you to select a line and the quadruple click a section (paragraph).

2. Pressing Ctrl + A.

Ctrl+A | Ctrl+E: Select all. Depending on the programs and the translations, it will be one or the other.

3. Double click within the word.

In most text editors, a double-click on a word selects it in its entirety.

4. Pressing key combinations Shift + arrow key.

If you press one of the arrow keys with the Shift key pressed, you can select a section of text.

5. Pressing key combinations Ctrl + End.

Ctrl + End: Go to the end of the document or window.

Ctrl + Home: Go to the beginning of the document or window.

Ctrl + Insert: Copy the selected text.

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7 0
2 years ago
Read 2 more answers
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms
inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

S = 1.23*10^9 W/m^2

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) 6.81*10^5 N/c

b) 2.27*10^3 T

Explanation:

To find the RMS value of the electric field, let's use the formula:

E_r_m_s = sqrt*(S / CE_o)

Where

C = 3.00 * 10^-^8 m/s;

E_o = 8.85*10^-^1^2 C^2/N.m^2;

S = 1.23*10^9 W/m^2

Therefore

E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}

E_r_m_s= 6.81 *10^5N/c

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

B_r_m_s = E_r_m_s / C;

= 6.81*10^5 N/c / 3*10^8m/s;

B_r_m_s = 2.27*10^3 T

8 0
4 years ago
Read 2 more answers
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

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2 years ago
Define electric potential energy
Verizon [17]
Electric potential energy, or Electrostatic potential energy, is a potential energy that results from conservative Coulomb forces and is associated with the configuration of a particular set of point charges within a defined system.
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3 years ago
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What is the answer to 7Hm a dm
blagie [28]

Answer:

7 hectometers is 7000 decimeters

8 0
3 years ago
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