The options are;
A) The skater experiences acceleration both while skating in a straight line and while making her turn.
B) Because the skater does not change her direction or speed, she does not experience acceleration.
C) The skater only experiences acceleration when she is turning.
D) The skater only experiences acceleration while skating in a straight line.
Answer:
A: The skater experiences acceleration both while skating in a straight line and while making her turn
Explanation:
We are told that She speeds up on the straight portion of the rink. This means that she experienced an acceleration since the speed was not constant but gradually increasing.
Also, we are told that She slows down near each end of the rink and then turns around. This means that she reduced speed before turning and it means there was also an acceleration as well when making the turn.
Thus, she experience acceleration both in the straight line and when making turn.
Option A is correct
Answer:
part 1 is from the article
Explanation:
Answer:

Explanation:
Molarity is a measure of concentration in moles per liter.

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.
- molarity= 1.2 mol/L
- moles of solute= 4.0 mol
- liters of solution =x
Substitute the values into the formula.

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.



x is being multiplied by 1.2 moles per liter. The inverse of multiplication is division, so divide both sides by 1.2 mol/L


The units of moles (mol) will cancel.


The original measurements both have 2 significant figures, so our answer must have the same. For the number we found, this is the tenths place.
The 3 in the hundredth place tells us to leave the 3 in the tenths place.

Approximately <u>3.3 liters of solution</u> are needed to make a 1.2 M solution with 4.0 moles of sodium chloride.