Answer:
the time taken t is 9.25 minutes
Explanation:
Given the data in the question;
The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V
now, every minute, the charge lost is 9.9 %
so we need to find the time for which the charge drops below 800 mV or 0.8 V
to get the time, we can use the formula for compound interest in basic mathematics;
A = P × ( (1 - r/100 )ⁿ
where A IS 0.8, P is 2.1, r is 9.9
so we substitute
0.8 = 2.1 × ( 1 - 0.099 )ⁿ
0.8/2.1 = 0.901ⁿ
0.901ⁿ = 0.381
n = 9.25 minutes
Therefore, the time taken t is 9.25 minutes
I believe C is your answer (also, I am also using tht same exact program!)
<h2>0.054×2.33×90</h2><h3>=0.11582×90</h3><h3>=11.3238</h3>
please mark this answer as brainlist
Answer:
Since it is falling freely, the only force on it is its weight, w.
w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
