Answer:
The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Explanation:
This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Therefore this is the formula for Half wave rectifier
Vrms = Vm/2 and Vdc
= Vm/π:
Where,
Vrms = rms value of input
Vdc = Average value of input
Vm = peak value of output
Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.
Answer:
The final velocity of the rocket is 450 m/s.
Explanation:
Given;
initial velocity of the rocket, u = 0
constant upward acceleration of the rocket, a = 18 m/s²
time of motion of the rocket, t = 25 s
The final velocity of the rocket is calculated with the following kinematic equation;
v = u + at
where;
v is the final velocity of the rocket after 25 s
Substitute the given values in the equation above;
v = 0 + 18 x 25
v = 450 m/s
Therefore, the final velocity of the rocket is 450 m/s.
Answer:
Timing Diagrams 15 pts. A 10 MHz clock that generates a 0 to 5V pulse train with a 30% duty cycle is connected to input X of a two input OR gate that has a 20nS propagation delay. The clock also goes to an inverter with a 10 ns propagation delay. The output of the inverter goes to the Y input of the OR gate. a) Draw the circuit. 2 pts. b) Plot the output of the clock for two cycles. Show times and voltages. 5 pts. c) On the same page as part (b) plot the output of the inverter. Show times and voltages. 3 pts. d) On the same page as parts (b & c) plot the output of the OR gate. Show times and voltages. 5 pts.
Answer:
The answer is below
Explanation:
a) The work done is equal to the loss in kinetic energy (KE)
Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy
Final KE = 
But the final velocity is 0 (at rest). Hence:
Final KE = 
ΔKE = 0 - K = -K
W = ΔKE = -K
Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)
W = qEd
but q = -e, hence:
W = -e * E * d
Using:
W = ΔKE
-e * E * d = -K
E= K / (e * d)
b) The electric field is in the direction of the electrons motion
False because it’s the same size all around!
