The radius of the earth is about 4,000 miles. If there were a global superhighway that circles around the entire equator of the
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Answer:
Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram
Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely
used integrated circuits for creating clock pulses is called a 555 timer. shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?
ANSWER : R1 = 144.3Ω, R2 = 72.2 Ω
Explanation:
Frequency = 10 MHz
Time period = 1 / F = 0.1 <em>u </em>s
Duty cycle = 75% = 0.75
Duty cycle can be represented as : Ton / T
Also: Ton = Th = 0.75 * 0.1 <em>u </em>s = 75 <em>n</em> s
TL = T - Th = 100 <em>n</em>s - 75 <em>n</em> s = 25 <em>n</em> s
To find the value of R2 we use the equation for time spent in the low (0 V) state
TL = R2*C*ln(2)
hence R2 = TL / ( C * In 2 )
c = 500 pF
Hence R2 = 25 / ( 500 pF * 0.693 ) = 72.2 Ω
To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,
Th = (R1 + R2)*C*ln(2)
from the equation make R1 the subject of the formula
R1 = (Th - ( R2 * C * In2 )) / (C * In 2)
R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )
R1 = ( 75 ns - ( 25 ns ) / 500 pf * 0.693
= 144.3Ω
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV