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scoundrel [369]
3 years ago
6

An advanced computer sends information to its various parts via infrared light pulses traveling through silicon fibers (n = 3.50

). To acquire data from memory, the central processing unit sends a light-pulse request to the memory unit. The memory unit processes the request, then sends a data pulse back to the central processing unit. The memory unit takes 0.50 ns to process a request. If the information has to be obtained from memory in 2.00 ns, what is the maximum distance the memory unit can be from the central processing unit?
Physics
1 answer:
sasho [114]3 years ago
8 0

Answer:

d = 6.43 cm

Explanation:

Given:

- Speed resistance coefficient in silicon n = 3.50

- Memory takes processing time t_p = 0.50 ns

- Information is to be obtained within T = 2.0 ns

Find:

- What is the maximum distance the memory unit can be from the central processing unit?

Solution:

- The amount of time taken for information pulse to travel to memory unit:

                            t_m = T - t_p

                            t_m = 2.0 - 0.5 = 1.5 ns

- We will use a basic relationship for distance traveled with respect to speed of light and time:

                           d = V*t_m

- Where speed of light in silicon medium is given by:

                           V = c / n

- Hence,              d = c*t_m / n

-Evaluate:           d = 3*10^8*1.5*10^-9 / 3.50

                           d = 0.129 m 12.9 cm

- The above is the distance for pulse going to and fro the memory and central unit. So the distance between the two is actually d / 2 = 6.43 cm

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tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
4 years ago
If the distance between two masses is tripled, the gravitational force between changes by a factor of
maw [93]

A. 1/9

Explanation:

The gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the two masses

r is the distance between the two masses

From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

\frac{1}{r^2}=\frac{1}{3^2}=\frac{1}{9}

6 0
4 years ago
Read 2 more answers
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
3 years ago
008 (part 3 of 4) 3.0 points
motikmotik

Car A take a time of 2.55hr and car B take a time of 2.14 hr

We know that distance divide by time is speed

here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr​

so speed=distance/time

s=d/t

t=d/s

=189/74

=2.55hr

In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr  

s=d/t

t=d/s

=199.8/93

=2.14hr  

so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h

learn more about Speed here brainly.com/question/13943409

#SPJ9

5 0
1 year ago
PLEASE SOLVE FAST!!! If the average American watches hours of TV every day , how many minutes will be spent in front of the TV b
masha68 [24]

Answer:

5694000 min

Explanation:

Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.

(4 h/day) × (60 min/1 h) = 240 min/day

They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.

240 min/day × (365 day/year) = 87600 min/year

They watch 87600 min/year. Finally, let's calculate how many minutes they spend watching TV in 65 years.

87600 min/year × 65 year = 5694000 min

6 0
4 years ago
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