The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
where q is the charge and
is the potential difference.
In our problem, the work done is W=39 J while the potential difference of the battery is
, so we can find the charge transferred by the battery:
The answer is c you got to look for answers that make sense
Answer:
350 F to 100 F it take approx 87.33 min
Explanation:
given data
oven = 350◦F
cooling rack = 70◦F
time = 30 min
cake = 200◦F
solution
we apply here Newtons law of cooling
= -k(T-Ta)
= 
(T(t) -Ta)
= 
= -k(T-Ta)
-ky = -ky
T(t) -Ta = (To -Ta) 
T(t) = Ta+ (To -Ta)
put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F
so here
200 = 70 + ( 350 - 70 ) 
k = 0.025575
so here for T(t) = 100F
100 = 70 + ( 350 - 70 ) 
time = 87.33 min
so here 350 F to 100 F it take approx 87.33 min
DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
Answer:POGGERS thx i rl needed this have a good day also something a regret the most is searching up henhai i am now addicted
