Question

The tension caused by a wave moving along a string is found using the formula
T=mv^2/L
If m is the mass of the string in grams, L is the length of the string in centimeters, and v is the velocity of the wave in centimeters per second, what is the unit of the tension of the string, T

Answer 1

Are you missing some data here?

Answer 2

no that is the question exactly from the paper. its my first question so i really need help

Answer 3

@laura022399 I can help you.

If I tell you that a car's path is described by this formula:

d=rt

and that d is distance, r is the speed of the car measured in miles per hour, and t is the time measured in hours, what is the unit of distance? \[\LARGE d=rt \\ \LARGE [d]=\frac{miles}{hour}hour\]
I put brackets around the d to show that this isn't the value, it's just the units. Since we have hours divided by hours, they cancel out to make 1 just like anything else. \[\LARGE [d] = miles\] This should make sense. Now apply this same reasoning to your problem Just plug in the units to the equation and see what units you get out on the other side. Try it out and I'll help if you make a mistake don't worry.

Answer 4

T=grams(centimeters per second)/ centimeters
And the centimeters cancel out so T=grams per second?
@Kainui

Answer 5

Almost, except remember, that velocity is squared.

Answer 6

OK so T=grams per second^2

Answer 7

Better, but still not quite right. Start over and see if you can catch your mistake, you didn't fully square everything.

Answer 8

T= grams( centimeters per second)^2/ centimeters
Is that correct like the starting equation is that right?

Answer 9

yes perfect so far! =)

Answer 10

T=grams(seconds)^2
Would this be how the ending is written?? Like would this be my final answer??

Answer 11

Nope, not quite, I think what you're missing is this step: \[\LARGE (\frac{centimeters}{second})^2=\frac{centimeters^2}{second^2}\]

Answer 12

What I am getting is that it T=grams^2

Answer 13

Wait a second, now you're going in the wrong direction. All I'm saying is this isn't right: \[\frac{grams \ centimeter^2}{second^2centimeter}=\frac{grams \ centimeter}{second^2}\]

Answer 14

Ok wait why is seconds^2 on the botton not the top

Answer 15

Because seconds are on the bottom in velocity.

Answer 16

OHHH ok! so T=grams/seconds^2

Answer 17

Nope, the centimeters in velocity is squared and there is only one centimeter from L in the bottom to cancel out. So there's still a centimeter in the top part.

Answer 18

so T=grams centimeters/ second^2

Answer 19

Yep, exactly. Think about it and make sure you understand why that's true. You should be able to work it out.

Answer 20

OMG everything just clicked!! OMG thank you so much!! ahh youre really nice! Thank you once again!!

Answer 21

Haha no problem, glad I could help!