The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig
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Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The power created is
Explanation:
From the question we are told that
The that the average power is mathematically represented as
Where W is is the Workdone which is mathematically represented as
Where F is the applies force and s is the displacement due to the force
So
Now this displacement can be represented mathematically as
Where is the average velocity and
is the time taken
So
=>
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>
<span>And find Tom's acceleration:
</span>
<span>
</span>
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>
<span>And find Tom's acceleration:
</span>
</span>
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
