We have the equation of motion , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.
Here s = 300 m, u = 0 m/s, a = 9.81
Substituting
Now we have v = u+at, where v is the final velocity
Here u = 0 m/s, a= 9.81 and t = 7.82 seconds
Substituting
v = 0+9.8*7.82 = 76.68 m/s
The speed with which the penny strikes the ground = 76.68 m/s.
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