Ask question

Ask question

All categories

2 years ago

8

A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the pro

cessor every 20 ms. The interface of the device includes two ports: one for status and one for data output. How long does it take to scan and service the device, given a clocking rate of 8 MHz? Assume for simplicity that all pertinent instruction cycles take 12 clock cycles

1 answer:

Lerok [7]2 years ago

6 0

Answer:

0.0000045 s

Explanation:

f = Frequency = 8 MHz

Clock cycle is given by

$\dfrac{1}{f}=\dfrac{1}{8\times 10^6}=1.25\times 10^{-7}\ s$

Time taken for 12 clock cycles

$12\times 1.25\times 10^{-7}=0.0000015\ s$

Time taken per instruction is 0.0000015 s

In reading and displaying information it requires 3 processes

1 for reading, 1 for searching and 1 for displaying.

$3\times 0.0000015=0.0000045\ s$

Time taken is 0.0000045 s

You might be interested in

Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc

lubasha [3.4K]

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

8 0

2 years ago

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

2 years ago

Two billiard balls of equal mass are traveling straight toward each other with the same speed. They meet head-on in an elastic c

Rus_ich [418]

Answer:

0 kg m/s before and after collision

Explanation:

Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:

P = mv - mv = 0 kg m/s

As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.

5 0

2 years ago

A compact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible fri

enot [183]

(a) 80 N/m

The spring constant can be found by using Hooke's law:

$F=kx$

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

$k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m$

(b) 0.84 Hz

The frequency of oscillation of the system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz$

(c) 1.05 m/s

The maximum speed of a spring-mass system is given by

$v=\omega A$

where

$\omega$ is the angular frequency

A is the amplitude of the motion

For this system, we have

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

$A=0.200 m$ (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

$v=(5.25 rad/s)(0.200 m)=1.05 m/s$

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

$a=\omega^2 A$

Using the numbers we calculated in part c):

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

$A=0.200 m$

we find immediately the maximum acceleration:

$a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2$

(f) At the position of maximum displacement: $x=\pm 0.200 m$

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

$a=\frac{F}{m}$

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

$F=kx$

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

$E=K=\frac{1}{2}mv_{max}^2$

where

m = 2.90 kg is the mass

$v_{max}=1.05 m/s$ is the maximum speed

Solving for E, we find

$E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J$

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

so the elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J$

and since the total energy E = 1.60 J is conserved, the kinetic energy is

$K=E-U=1.60 J-0.18 J=1.42 J$

And from the relationship between kinetic energy and speed, we can find the speed of the system:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s$

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

So the restoring force exerted by the spring on the mass is

$F=kx=(80 N/m)(0.0667 m)=5.34 N$

And so, we can calculate the acceleration by using Newton's second law:

$a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2$

8 0

2 years ago

Para proteger un computador de sobrecargas eléctricas, Juan coloca un filamento delgado de cobre llamado fusible en su circuito,

Lynna [10]

Answer:

Los fusibles están diseñados de tal forma que estos se "rompen" o se funden, cuando la demanda eléctrica supera un dado valor (cuando demasiada electricidad pasa a través de el).

Una vez el filamento se rompe, la corriente ya no puede circular por el (podes pensar en esta situación como un cable roto, la electricidad no puede circular por este cable)

Entonces, al romperse el filamento, en caso de una sobrecarga eléctrica, el flujo de electricidad se corta, y de esta forma se protege al computador de posibles sobrecargas.

7 0

2 years ago