Recent content by Callmelucky

yeah, but I was confused why the method I was using so far isn't working anymore. Thank you.

Okay, thank you very much. :D

Ohh, that make sense, should I always do that when moving 2 down? Because I haven't done that once so far

I didn't learn about domain, in Croatia we learn that in 4th grade of high school, I am in 2nd right now. Are you saying that I can't move 2 down if it's original place is up?

So basically this is how I solved this problem: 1. ##f(x)=\log _{2} x^2 - 1## 2. ##0=\log _{2} x^2 -1 ## 3. ##1= 2\times \log _{2} x## 4. ##\frac{1}{2}= \log _{2} x## 5. ##2^{\frac{1}{2}}=x=\sqrt{2}## So I wrote coordinates to be (##\sqrt{2}##, 0) But apparently, that is not the only solution...

it's actually a workbook by Nada Brkovic, there is a question(273) if the friction coefficient can be greater than 1, and, somehow I concluded that it can not be greater than 1 if the body is moving. The best thing is that it is not a proper question, it's more like a side question along with 3...

@MatinSAR, @haruspex. I got it, thank you. @kuruman, I had no idea that mi can be >=1, there was a question in the textbook if mi can be >= 1 and the answer at the end was that it can not. I never checked that, just accepted it as a fact, but after checking I realized that it obviously can be...

So basically I need to find the coefficient of friction given the listed information. What bothers me is that I am getting two different accelerations for two different approaches. When I calculate acceleration using Fg=mgsin60 I do it this way: Fg=mgsin60 -> ma=mgsin60 ->a=gsin60 -> a=8.66. But...

thank you

thank you

I wonder if it's ##f(x)=2^{x}-1## considered an exponential function because in my textbook it's stated that the set of values of an exponential function is a set of positive real numbers, while when graphing this function I get values(y line) that are not positive(graph in attachments), so I am...

Does that then mean that the body didn't cross any distance for that one second? One more question, I have calculated that the average speed in the first 6 seconds is 3/2 m/s but the solution in the textbook is 10/6 m/s. I used the same formula for distance traveled by accelerating the body for...

my textbook, it's just marked as s instead of d

So basically I wonder why the distance traveled by a body in the 5th second gives different results when calculated by the formula for accelerating body(##d=V_0\times t + \frac{1}{2}\times at^{2}##) and when calculated using a graph(formula for the surface of the triangle). Here is the graph of...

I found my mistake. What an idiot I am. I plugged the value of (a) instead of (b), and instead of multiplying with 0.8 I divided it by 0.8, therefore got the wrong result. I am sorry for waisting everybody's time. Thank you.