Mathematics: Atemperature of 221/3 degrees below zero

Atemperature of 221/3 degrees below zero - id22132996448, jaxheisjax532

Arealbot

15.02.2022

6.82 mt^3/min

Step-by-step explanation:

By congruence of triangles, the radius r of the cone base when its height is 1 meter satisfies the relation

r/1 = 2.5/6

(See picture attached)

So, r = 0.4166 meters when the water is 1 meter high.

The volume of a cone with radius of the base = R is given by

$V=\frac{\pi R^2h}{3}$

So, the volume of water when it is 1 meter high is

$V=\frac{\pi* 0.(4166)^2}{3}=0.1817\;mt^3$

One minute later, the height of the water is 1 meter + 30 centimeters = 1.30 mt

The radius now satisfies

r/1.3 = 2.5/6

and now the radius of the base is

r = 0.5416 mt

and the new volume of water is $0.3071\;mt^3$

So, the water is raising at a speed of

0.3071-0.1871 = 0.12 mt^3/min

This speed equals the difference between the water being pumped and the water leaking,

So,

0.12 = Speed of water being pumped -6.8

and

Speed of water being pumped = 6.8+0.12 = 6.82 mt^3/min

0,0(0 оценок)