Hello ? would someone be able to tell me if I have solved this question correctly? The last part of it 4(c) makes me think I must be missing something but I’m just not sure. Thanks in advance ..
I presume you are trying to calculate the magnitude in part c). The magnitude of a vector is [imath]|| \textbf{v} || = \sqrt{ \textbf{v} \cdot \textbf{v} }[/imath],Hello ? would someone be able to tell me if I have solved this question correctly? The last part of it 4(c) makes me think I must be missing something but I’m just not sure. Thanks in advance ..
Actually, in Physics [imath]\textbf{v} \cdot \textbf{v}[/imath] is often denoted as [imath]v^2[/imath]. As long as you have context you should be fine.y*y is not y^2
xy is not x*y
Use correct math symbols.
Don't tell Steven about Physics - it is too uncertain for him.Actually, in Physics [imath]\textbf{v} \cdot \textbf{v}[/imath] is often denoted as [imath]v^2[/imath]. As long as you have context you should be fine.
-Dan
Hey, stop talking about me!Don't tell Steven about Physics - it is too uncertain for him.
You forgot to take the square root at the end.Thanks for all the help! Okay .. I think I got it .. Is this correct?