HELP MEEEE FINALS THIS WEEK!!! I’ve gotten those so far I just need the other ones!!

Sally and Sue took a canoe trip down the Mississippi River. Their average speed for the trip south was 20 miles/hour. 20 miles/h

Anna35 [415]

C

Speed is the time rate at which Sue covers the distance. It is derived by dividing the total distance covered by the total time taken to cover the distance . Usually SI unit for speed is km/h or mph or m/s.

Explanation:

Speed is also referred to as velocity- so the two are synonymous.

Acceleration is the rate at which speed is increasing. It is usually given by SI unit m/s². The opposite of acceleration is deceleration which is the rate at which speed is decreasing.

Distance is the measurement, in meters of kilometers or miles or yards..etc, that has been covered from one point to another.

Learn More:

For more on speed, velocity, distance and acceleration check out;

brainly.com/question/13874410

brainly.com/question/1822168

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6 0

3 years ago

Read 2 more answers

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.

6 0

3 years ago

Sodium sulphate reacts with calcium nitrate to produce sodium nitrate and calcium sulphate. A) identify the reactants and produc

Rudik [331]

Answer:

yes ...................

6 0

3 years ago

A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature

Aliun [14]

Answer:

"1.4 mL" is the appropriate solution.