Question

Found part 1, troubled by part 2.      

In 2006, NASA’s Mars Odyssey orbiter detected violent gas eruptions on Mars, where the acceleration due to gravity is 3.7 m/s2. The jets throw sand and dust about 79.0 m above the surface.

Answer 1

What is the speed of the material just as it leaves the surface?

I found that speed to be: |v| = \(\color{red}{24.161~\rm m/s}\)
(this is correct)

Answer 2

and this is where I am having an issue: (part 2)

Scientists estimate that the jets originate as high-pressure gas speeds through vents just underground at about 160 km/h. How much energy per kilogram of material is lost due to nonconservative forces as the high-speed matter forces its way to the surface and into the air?

Answer 3

Oh, so you just apply the formula for kinetic energy (0.5mv^2), with the force of this gas?

Answer 4

But, where did you get 23.5584 ?

Answer 5

that was the answer for part a but your is just little different than mine

Answer 6

So, basically ?

\(KE=\frac{1}{2}m(\Delta v)^2=\frac{1}{2}m(v_f-v_o)^2\)

Answer 7

mhm

Answer 8

what does that stand for?

Answer 9

what?

Answer 10

mhm

Answer 11

are you talking about mhm? tht means yea

Answer 12

Oh, ok.

Answer 13

I'm not good with American slang.

Answer 14

lol not really a slang you know when people go like hmm so same thing

Answer 15

Alright.

Answer 16

I used my number for speed,
24.161

and I got 205.71 J/kg

Answer 17

wow, that was wrong.
I guess your number was more precise.

Answer 18

i mean idk but thats what i came up with lol

Answer 19

218.11

Answer 20

and that is also wrong

Answer 21

ohh really? what was the right number?

Answer 22

Will see if I can try one more time to get it right.

Answer 23

alright. i gtg now.

Answer 24

E (Joules) = 1/2 × (m kg) × (v m/s)^2
E (Joules) ÷ (m kg) = 1/2 × (v m/s)^2

So maybe,

E (Joules) ÷ (m kg) = 1/2 × (\(v_f\) m/s)^2 - 1/2 × (\(v_i\) m/s)^2

Answer 25

Yup!

710 J/kg

Answer 26

it is the change in energy, not a change in velocity ...

We did,
E (Joules) ÷ (m kg) = 1/2 × (\(v_f-v_i\) m/s)^2

when should have done
E (Joules) ÷ (m kg) = 1/2 × (\(v_f\) m/s)^2 - 1/2 × (\(v_i\) m/s)^2

Answer 27

we find the gain -710 J/kg,
so the loss is 710 (positive) J/kg