Advertisements
Advertisements
Question
Make k the subject of formula T = `2pisqrt(("k"^2 + "h"^2)/"hg"`
Advertisements
Solution
T = `2pisqrt(("k"^2 + "h"^2)/"hg"`
⇒ `"T"/(2pi) = sqrt(("k"^2 + "h"^2)/"hg"`
Squaring both sides
⇒ `("T"/(2pi))^2 = ("k"^2 + "h"^2)/"hg"`
⇒ `"hg"("T"/(2pi))^2 - "h"^2` = k2
⇒ k = `sqrt(("T"^2"hg")/(4pi^2) - "h"^2`.
APPEARS IN
RELATED QUESTIONS
How many minutes are there in x hours, y minutes and z seconds.
Make L the subject of formula T = `2pisqrt("L"/"G")`
Make a the subject of formula S = `("a"("r"^"n" - 1))/("r" - 1)`
Make r2 the subject of formula `(1)/"R" = (1)/"r"_1 + (1)/"r"_2`
Make c the subject of formula x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"a")`
If V = pr2h and S = 2pr2 + 2prh, then express V in terms of S, p and r.
Make a the subject of the formula S = `"n"/(2){2"a" + ("n" - 1)"d"}`. Find a when S = 50, n = 10 and d = 2.
Make I the subject of the following M = `"L" /"F"(1/2"N" - "C") xx "I"`. Find I, If M = 44, L = 20, F = 15, N = 50 and C = 13.
"The volume of a cone V is equal to the product of one third of π and square of radius r of the base and the height h". Express this statement as a formula. Make r the subject formula. Find r, when V = 1232cm3, π = `(22)/(7)`, h = 24cm.
"Area A oof a circular ring formed by 2 concentric circles is equal to the product of pie and the difference of the square of the bigger radius R and the square of the bigger radius R and the square of the smaller radius r. Express the above statement as a formula. Make r the subject of the formula and find r, when A = 88 sq cm and R = 8cm.
