It is a correction for the answer to the normal component in "Find the tangential & the normal component of..."?
Correction: it should be that c=|r'xr"|/|r'^3|=>c=6t^2/[t^3(4+9t^2)^1.5]=>c=6/[t(4+9t^2)^1.5]. Thus, cv^2=6(4+9t^2)t/(4+9t^2)^1.5=>cv^2=6t/sqr(4+9t^2)=
the normal component. Sorry.Mathematics11 months ago
- 1 AnswerMathematics2 years ago
- 4 AnswersMathematics4 years ago
- 2 AnswersMathematics7 years ago
A ladder of 4 m leans against a wall with its feet on the floor. Between the ladder & the wall there is a cubic box of length 1 m at the foot of the wall & the inclined ladder is just touching a corner of the box.
Find the possible heights of the ladder from ground. (Round to 2 decimal places).3 AnswersMathematics7 years ago
calculus nor mathematical induction.
guys please.2 AnswersMathematics7 years ago