Question

An automobile engineer is redesigning a conical chamber that was originally specified to be 12 inches long with a circular base of diameter 5.7 inches. In the new design, the chamber is scaled by a factor of 1.5.

Answer 1

The volume of the original chamber is 102.02 248.04 306.06 412.08 cubic inches. This is 242.47 344.49 432.16 448.21 cubic inches less than the volume of the new chamber

Answer 2

those are the choices

Answer 3

Do you know the formula for the volume of a cone?

Answer 4

\[V_{cone} = \pi r^2 \frac{ h }{ 3 }\]
Where r = radius
and h = height

Answer 5

Let me know how you would approach this problem

Answer 6

ik that

Answer 7

idk how to do it

Answer 8

Solve for the volume of the first cone, then multiply it by the factor of 1.5 to get the new cone.

Answer 9

i got the first one now what is the second one

Answer 10

@shadow

Answer 11

More accurately, multiply the measurements by 1.5 to get the new cone

Answer 12

i didn't get any of the answer choice

Answer 13

Let me work it out

Answer 14

kk

Answer 15

did u get the answer

Answer 16

\[V_{cone} = \pi r^2 \frac{ h }{ 3 }\]
r = d/2
r = 5.7/2
r = 2.85
\[V_{1} = \pi r^{2}_{1} \frac{ h_{1} }{ 3 }\]
r1 = 2.85
h1 = 12
\[V_{1} = \pi (2.85^{2})( \frac{ 12 }{ 3 })\]
\[V_{1} = \pi(8.1225)(4) = 102.1\]

Close enough to 102.02, they probably rounded something differently.

\[V_{2} = \pi r^{2}_{2} \frac{ h_{2} }{ 3 })\]
\[r_{2} = r_{1} \times 1.5\]
\[r_{2} = 2.85 \times 1.5 = 4.275\]
\[h_{2} = h_{1} \times 1.5 \]
\[h_{2} = 12 \times 1.5 = 18\]
\[V_{2} = \pi (4.275)^2 (\frac{ 18}{ 3 })\]
\[V_{2} = \pi (18.275625)(6) = 344.49\]

Answer 17

Legend:
V1 = volume of cone 1
r1 = radius of cone 1
h1 = height of cone 1

I am sure you get the idea. Same concept applies for the second cone. I did this so you know how to properly account for increasing the cone by a scale factor of 1.5.

Answer 18

Moved to the Mathematics section.