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dalvyx [7]
3 years ago
14

Suppose we want to determine how many of the bits in a twelve-bit unsigned number are equal to zero. Implement the simplest circ

uit that can accomplish this task.

Engineering
1 answer:
timurjin [86]3 years ago
8 0

Answer:

See the picture for circuit figure.

Explanation:

First load 12 bit unsigned number on shift register and apply clock tick to increment counter from LSB bit.

Logic 1 is shifted into register to ensure counter doesn't count more than 12 bits.

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An inductor is energized as in the circuit of Fig. 2-4a. The circuit has L 10 mH and VCC 14 V. (a) Determine the required on tim
Savatey [412]

Answer:

A) 11.1 ms

B) 5.62 Ω

Explanation:

L ( inductance ) = 10 mH

Vcc = 14V

<u>A) determine the required on time of the switch such that the peak energy stored in the inductor is 1.2J </u>

first calculate for the current  ( i )  using the equation for energy stored in an inductor hence

i = \sqrt{\frac{2W}{L} }   ----- ( 1 )

where : W = 1.2j ,  L = 10 mH

Input values into equation 1  

i = 15.49 A

Now determine the time required  with expression below

i( t ) = 15.49 A

L = 10 mH, Vcc = 14

hence the time required ( T-on ) = 11.1 ms

attached below is detailed solution

B) <u>select the value of R such that switching cycle can be repeated every 20 ms </u>

using the expression below

τ = \frac{L}{R}  ---- ( 2 )

but first we will determine the value of τ

τ = t-off / 5 time constants

  = (20 - 11.1 ) / 5  = 1.78 ms

Back to equation 2

R = L / τ

  = (10 * 10^-3) / (1.78 * 10^-3)

  = 5.62 Ω

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The voltage valve at which a zirconia O2S switches from rich to lean and lean to rich is
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If aligned and continuous carbon fibers with a diameter of 9.90 micron are embedded within an epoxy, such that the bond strength
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The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
3 years ago
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