@Nathanael Donald Don't fall for this, It's a scam

Yeah Columbine does IB as well

Instablaster

ong school in america is so easy i wishhhh i could live there

@@lesdeliresdecharlotte6440 where do you live good sir? france?

😂😂😂lying

IB physics for me. Its damn confusing for the first couple days

Mans was glitching 😹

You're right: I messed up carrying that onto the next slide. I just added a note. Thanks for the edit!

Marco Polo ji

you divide your value by 1000

You must multiply your answer by 1000, for 1 kg. is 1000 g., and 1 m. is 100 cm. When you plug in the corresponding values, you get the following: (1/100)/(1/100)^3, which is (10^6)/(10^3), which is 1000. Therefore, you multiply your g/cm^3 value by a factor of 1000 to convert to kg/m^3. Q.E.D.

bit late but its just how it works

You're left with a challenge here. Without uncertainties reported with the measurements, or knowledge of how the measurements were made (caliper? ruler? what units?) you are left with making assumptions. The initial measurements were missing information; no amount of calculation can fix that. There is no real way to give a great answer here. If we make some guesses, we do know that the last digit reported is an estimate, and often the uncertainty is half of the smallest decimal value reported. Since each value has a hundredths decimal, a possible absolute uncertainty would be + or - .005 . This is a big guess, though. If you measure with a ruler, you are matching two ends of a ruler (the zero point and the end point) up with your object. That is two error possibilities, so you would need to double your uncertainty. If you used calipers, only one measurement was made. See where we are stuck? When we start guessing or assuming, we undermine our work. Anyone else have helpful thoughts?

@@ckillmerportageps apologies for not fully specifying the question... the measurement were made with a vernier caliper

@@pickleporkdork7358 If you took the measurements, what were the smallest divisions on the caliper? Did it show to the tenth place? If so, I recommend using + or - .05 as your uncertainty Your caliper may have come with documentation that indicates it has a narrower range than that, however.

I ran across this article. web.uvic.ca/~jalexndr/192UncertRules.pdf For square rooting, treat it as though you are raising the uncertainty by the power of 1/2 (for example, the square root of 4 is the same thing as 4^(1/2)). Then, you will multiply the relative uncertainty by the power (in this case, 1/2). Example: ( 9g plus or minus 1%)^1/2 = (3g plus or minus .5%)

Colin Killmer Thank you so much. Also, how to find out uncertainities while operating with logarithms?

Because division is the same as multiplication of the inverse of the number.

0.1 % + 0.02 % 0.12

And this is what I was looking for. Thanks.

You guy are funny, instead saying wrong,go head and help solve it yourself, this will helpfull all student ms Harvard. Thanks

Lanoline, you are right. It is a shame that such videos go unchallenged. They are hurting students.

your second proposal is correct...the first one is not : if you round 0.0025 it is 0.003..not 0.03

@@skatiaris21 False. The precision of the answer must be the same than the precision of the uncertainty.

Mike Astro must the same of what? retype correctly please

Mike Astro i guess u say the same number of decimal places...thats true but you cant transform 0.003 to 0.03...u multiply a number by 10 arbitrarily...

@@skatiaris21 You are right. It is 0,003 and not 0,03. So, in the uncertainty is 0,003, the 4,45 must become 4,450 because the uncertainty must be at the same level that the measure. For the rounding, there is a rule (rounding at the even number) that says that if the number to round is even followed by a 5, we let the even number like that. So, 0,035 would become 0,04, but 0,025 would become 0,02.

I had this law but I don’t know how to use it : (delta d) / d = (delta s) / s + (delta g)/g this example was from my mind , but the point is how do we use it ?? Thanks in advance