Solved Problem 03 | Cube

Problem 03
What is the weight of a block of ice 24 in. by 24 in. by 24 in., if ice weighs 92 per cent as much as water, and water weighs 62.5 lb per cu. ft.?
 

Solution 03
Unit weight of water
$\gamma_{water} = 62.5 \, \text{ lb/ft}^3$
 

Unit weight of ice
$\gamma_{ice} = 92\% \, \gamma_{water}$

$\gamma_{ice} = 0.92(62.5)$

$\gamma_{ice} = 57.5 \, \text{ lb/ft}^3$
 

Volume of ice block
$V_{ice} = (24/12)^3$

$V_{ice} = 8 \, \text{ ft}^3$
 

Weight of 8 ft3 ice block

Solved Problem 01 | Cube

Problem 01
Show that (a) the total surface of the cube is twice the square of its diagonal, (b) the volume of the cube is $\frac{1}{9}\sqrt{3}\,s^3$ times the cube of its diagonal.
 

Solution 01
Space diagonal $s = a\sqrt{3}$, thus, $a = \dfrac{s}{\sqrt{3}}$
 

(a) Show that A = 2s2
$A = 6a^2$

$A = 6\left( \dfrac{s}{\sqrt{3}} \right)^2$

$A = 6\left( \dfrac{s^2}{3} \right)$

$A = 2s^2$       okay!
 

Arithmetic, geometric, and harmonic progressions

Elements
a1 = value of the first term
am = value of any term after the first term but before the last term
an = value of the last term
n = total number of terms
m = mth term after the first but before nth
d = common difference of arithmetic progression
r = common ratio of geometric progression
S = sum
 

Arithmetic Progression, AP

Arithmetic progression is a sequence of numbers in which the difference of any two adjacent terms is constant. The constant difference is commonly known as common difference and is denoted by d. Examples of arithmetic progression are as follows:
 

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