Question

Someone wanna help me with geometry haha?

Answer 1

HMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM

Answer 2

HmhHMmhMHmhMHMMMMMMMMMM

Answer 3

What grade?

Answer 4

10

Answer 5

UHHHH, i can't then

Answer 6

kk

Answer 7

I mean what exactly do you need help with bc there are certain parts that I cant help in.

Answer 8

Just send the question or ss

Answer 9

how do i find density based on radius and length?

Answer 10

Oh yeah, I'm sorry, I can't help with that maybe someone else can, I hope you find what ur looking for tho.

Answer 11

thanks

Answer 12

what you got?

Answer 13

how do i find density based on radius and length?

Answer 14

so just you need to know the formula what you need using to calcule this density in function about what material you talk

Answer 15

send me a pic of it so i can read it pls

Answer 16

im trying i promise

Answer 17

kk thanks

Answer 18

Part B:

assuming the log has a perfect cylindrical shape, we can find its volume using \( V_{Log}=\pi r^2\times h \) where \( r=5in \) and \( h=30in \)

therefore:

\( V_{Log}=\pi 5^2\times 30 \)

\( V_{Log}=25\pi \times 30 \)

\( \boxed{V_{Log}=750\pi} \)

we know that density\( (\rho) = \dfrac{mass(m)}{volumme(v)} \)

But the thing says were given logs weight not it's mass, and to fid the mass we refer to \( weight(f) =mass(m)\times 9.806 m/s^2(g) \ \ \Rightarrow \ \ \boxed{mass(m)= \dfrac{ weight(f)}{9.806 m/s^2(g)} }\)

\( mass(m)=\dfrac{42.63\ lbs}{9.806m/s^2} \approx \boxed{4.347} \)

let's plug that back into finding the density:

\( \boxed{\rho= \dfrac{4.347}{750\pi}} \)

and that is our answer I believe.

Answer 19

Thank you so much!!

Answer 20

srry i tried

Answer 21

its all good!

Answer 22

ok