Question 46

[ladywait223593]

This problem is ridiculous. You're explanation should have been to guess if that's really how you expect us to solve this problem. If the solution can't be found by taking a ratio of the moon's gravity to earth's then solution is not to blame so much as the problem which is beyond likelihood of appearing on a real mcat. The simple fact is if you go off one decimal place or round any number then you wind up with an answer resembling none of your choices. Figure out a way to make this worth my time.

[ladywait223593]

the way i solve it gives me 2.3 s but the drawback is it requires having an approximation of the square root of 6. Maybe I'm not like most people because I didn't install a calculator chip in my brain at an early age but suffice it to say it's 2.5~. I'm not even going to go into the rest because as mentioned previously this problem is more trouble than it's worth.

[mcat_premed3832]

It is true that there are few problems on the real MCAT that requires detailed calculations. But it does happen occasionally and regularly. The calculation for this problem is not complex and estimates can be done without consequence:

L/g = .537 m / 1.67 = 1/3 (approx)

Tm = (2/root 3) pi

For the MCAT, you are expected to memorize root 3 (1.7) and root 2 (1.4):

Tm = 1.15 x 3.14 = 3.6

From the Table, we get the equivalent TE = 1.54

Tm - TE = 2.1 seconds

NB: the calculation can be made even faster if you know the results of the sides of a right angle triangle. You would recognize that the inverse of (root 3/2) which is .866. Thus you would have skipped one step above:

Tm = 3.14/.87 = 3.6