Hello!
The best explanation is the new "experimental evidence", which occur with the help of new and improved technology. For this question, I suggest you to answer letter b).
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The formula we can use in this case is:
d = v0t + 0.5 at^2
v = at + v0
where,
d = distance travelled
v0 = initial velocity = 0 since at rest
t = time travelled
a = acceleration
v = final velocity when it took off
a. d = 0 + 0.5 * 3 * 30^2
d = 1350 m
b. v = 3 * 30 + 0
<span>v = 90 m/s</span>
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - (
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V
In 16 times
KE= o.5 m times V squared