Why are these two answers wrong?
second one: must use this rule: \[\frac{ dV}{ dt } = \frac{ dV }{ dv }\frac{ dv }{ dt }\]
Hmm what numbers would I plug in though? would the dvs cancel?
find the derivative, so, (2.99*10^16)/r^2 which then is written as (2.99*10^16)*r^-2 then take the derivative like this... (2) (2.99*10^16)*r^-2 then this.. (2) (2.99*10^16)*r^-3 then convert it back to a fraction (2) (2.99*10^16) / r^3 thus 5.98x10^16 / r^3 then plug in r taking the derivative is needed because the phrase "rate of change" is used in the question.
For the second problem: V = -Blv ---- (1) v = 8t + 9 Substiture v in equation (1) Then find dV/dt
In first problem don't forget the negative sign when you find dF/dr.
DemolisionWolf why did you not do the derivative of the top? Also when I plugged in r I got 1.927237236e^-4
Ranga: 1=8t+9 -8=8t t=-1
No. v = 8t + 9 V = -Blv = -Bl(8t + 9) What is dV/dt = ?
so it would be -4.8(8t+9) and then would I distribute?
Yes, distribute and then find the derivative.
V=-38.4 ah thanks for help, do you know what went wrong with problem one?
Yes. First you have to find dF/dr when F(r) = (2.99 x 10^16)/r^2. DemolisionWolf has done it above. I will just add a negative sign to the result. Can you try finding dF/dr by yourself again?
Don't worry about the constant term initially. Just find derivative for 1/r^2 first.
F(r)=-5.98x10^16 / r^3 F(r)=1/2r?
so it would be flipped and -2r
OK, you got dF(r)/dr = -5.98x10^16 / r^3 which is correct. Just plug in r = 6.77 x 10^6
hmm I got -1.927237236e^-4, maybe I should use e instead of x10
You got mostly correct. But the e should be 10. The answer is -1.927237236 x 10^-4
I typed that in and it wants an exact answer so I x10^-4 and got -192723.724 but it still coming up as incorrect let me try again.
Try 0.00019272372
still no luck
Don't forget the negative sign
Does it say how many decimals in the answer or if the answer should be in scientific notation?
ahhh that got it lol, that weird why it did not take scientific notation, thanks a lot for the help!
Glad to help.
Join our real-time social learning platform and learn together with your friends!