Rates of Change (1 Viewer)

[12o9]

For some reason I cant get the right answer =/ could someone help ? tahnks in advance

 

[tommykins]

回复: Re: Rates of Change

dV/dt > 0 for evaporation

since evaporation reduces volume:
V(t) = V₀ - (1/400)(t - t^2/120)

i)
V(0) - V(30) = (1/400)(30 - 30^2/120)
= 0.05625 mL

ii)
evaporation ceases when dV/dt = 0
(1/400)(1 - t/60) = 0
60 = t

V(0) - V(60) = (1/400)(60 - 60^2/120)
= 0.075 mL would have evaporated

Knew part i) involved an integral, as dV/dt is rate of change.

 

[lyounamu]

(i) at t = 30,
dV/dt = 1/400(1-1/2)
= 1/800 mL
therefore 0.00125 mL will evaporate in the first 30 seconds.

(ii) Evaporation ceases when dV/dt = 0
1/400 - t/24000 = 0
t - 60 = 0
t = 60
therefore evaporation ceases at t = 60 seconds

The integral of dV/dt (V) = t/400 - t²/48000
at t = 60
V = 60/400 - 60²/48000
= 0.075 mL
therefore 0.075 mL of the solvent will have evaporated by 60 seconds

For i) you have to use an integration to find the volume evaporated in the first 30 seconds.

I think the rests are correct.

 

[12o9]

Awesome. thanks.

 

[Aerath]

dV/dt > 0 for evaporation

since evaporation reduces volume:
V(t) = V₀ - (1/400)(t - t^2/120)

i)
V(0) - V(30) = (1/400)(30 - 30^2/120)
= 0.05625 mL

When you integrate it - wouldn't you get a +c?

 

[trailblazer]

looks like a sbhs paper.. lemme guess, anthony?

 

[12o9]

looks like a sbhs paper.. lemme guess, anthony?

and you are ?

 

[hopethisworks]

and you are ?

LOL! sup anthony

 

[Aerath]

the C is V₀ and i just made the other term negative as evaporation reduces volume. hence the volume at any given time is the original volume V₀ minus the volume lost due to the rate of change

An - ok, thanks.

 

[Timothy.Siu]

An - ok, thanks.

lol noob