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3 years ago

An 80 kg long jumper in flight accelerates at a rate of 10 m/s^2. What is the force of the long jumper

Chemistry

1 answer:

Katarina [22]3 years ago

6 0

Answer:

F = 800 N

Explanation:

Given data:

Mass = 80 Kg

Acceleration = 10 m/s²

Force = ?

Solution:

Formula:

F = force

m = mass

a = acceleration

Now we will put the values in formula:

F = 80 kg <em>× </em>10 m/s²

F = 800 kg.m/s²

kg.m/s² = N

F = 800 N

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The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

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