Um relógio de ponteiros funciona durante um mês. Qual é, em N.C, o número de voltas, aproximadamente, que o ponteiro dos minutos
1 answer:
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Answer:
R = 9.85 ohm , r = 0.85 ohm
Explanation:
Let the two resistances by r and R.
when they are connected in series:
V = 12 V
i = 1.12 A
The equivalent resistance when they are connected in series is
Rs = r + R
So, By using Ohm's law
V = i Rs
Rs = V / i = 12 / 1.12 = 10.7 ohm
R + r = 10.7 ohm .... (1)
When they are connected in parallel:
V = 12 V
i = 9.39 A
The equivalent resistance when they are connected in parallel
So, By using Ohm's law
V = i Rp
Rp = V / i = 12 / 9.39 = 1.28 ohm
.... (2)
by substituting the value of R + r from equation (1) in equation (2), we get
r R = 8.36 ..... (3)
..... (4)
By solvng equation (1) and (4), we get
R = 9.85 ohm , r = 0.85 ohm
Answer:
(a) 1.939 m/h
(b) 0.926 m/h
(c) -0.315 m/h
(d) -1.21 m/h
Explanation:
Here, we have the water depth given by the function of time;
D(t) = 7 + 5·cos[0.503(t-6.75)]
Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;
D'(t) =
= 5×(-sin(0.503(t-6.75))×0.503
= -2.515×(-sin(0.503(t-6.75))
= -2.515×(-sin(0.503×t-3.395))
Therefore we have;
(a) at 5:00 AM = 5 - 0:00 = 5
D'(5) = -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h
(b) at 6:00 AM = 6 - 0:00 = 6
D'(5) = -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h
(c) at 7:00 AM = 7 - 0:00 = 7
D'(5) = -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h
(d) at Noon 12:00 PM = 12 - 0:00 = 12
D'(5) = -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.