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Mademuasel [1]
2 years ago
5

If the slowest instruction in the SCA executes in 12.5 ns, then what is maximum system clock frequency in MHz

Physics
1 answer:
Margaret [11]2 years ago
5 0

Since the slowest instruction in the SCA executes in 12.5 ns, the maximum system clock frequency is 80 MHz

To answer the question, we need to know what frequency is.

<h3>What is frequency?</h3>

Frequency is the number of oscillations per second of a wave.

It is given by f = 1/T where T = period of wave

Now, given that the slowest instruction in the SCA executes in t = 12.5 ns, we need to calculate maximum system clock frequency, f.

<h3>What is the maximum system clock frequency?</h3>

So, f = 1/t

= 1/12.5 ns

= 1/(12.5 × 10⁻⁹ s)

= 1/12.5 × 10⁹ Hz

= 0.08 × 10⁹ Hz

= 80 × 10⁻³ × 10⁹ Hz

= 80 × 10⁶ Hz

= 80 MHz

So, the maximum system clock frequency is 80 MHz

Learn more about maximum system clock frequency here:

brainly.com/question/14636488

#SPJ11

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2 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
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Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

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We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

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So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

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