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
Answer:
13.51 nm
Explanation:
To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians
y/L=tan θ ≈ θ
and ∆θ ≈∆y/L
Where ∆y= wavelength distance= 2.92 mm =0.00292m
L=screen distance= 2.40 m
=0.00292m/2.40m
=0.001217 rad
The grating spacing is d = (90000 lines/m)^−1
=1.11 × 10−5 m.
the small-angle
approx. Using difraction formula with m = 1 gives:
mλ = d sin θ ≈ dθ →
∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad
=0.000000001351m
= 13.51 nm
Answer:
Fault lines
Explanation:
Earthquakes are most likely to occur near or on fault lines. A great example of this is the ring of fire, a gigantic fault line that gives catastrophic earthquakes.
Answer: static stretching
Explanation:
e.g rubberband
Answer:
Explanation:
<u>Given Data:</u>
Mass = m = 4 kg
Acceleration due to gravity = g = 9.8 m/s²
Height = h = 1 m
<u>Required:</u>
Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = mgh
<u>Solution:</u>
P.E. = (4)(9.8)(1)
P.E. = 39.2 Joules
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
