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kogti [31]
3 years ago
5

Plot the function for . Notice that the function has two vertical asymptotes. Plot the function by dividing the domain of x into

three parts: one from –4 to near the left asymptote, one between the two asymptotes, and one from near the right asymptote to 4. Set the range of the y axis from –15 to 15.
Engineering
1 answer:
elena-s [515]3 years ago
8 0
This is a very very difficult one for me, let me get back to you with the proper answer.
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For the system in problem 4, suppose a main memory access requires 30ns, the page fault rate is .01%, it costs 12ms to access a
raketka [301]

Answer:

a. 7.75

b. 24.4

Explanation:

The Operating system uses virtual memory and page tables maps these virtual address to physical address. TLB works as a cache for such mapping.

program >>> TLB >>> cache >>> Ram

A program search for a page in TLB, if it doesn't find that page it's a TLB miss and then further looks for the page in cache.

If the page is not in cache then it's a cache miss and further looks for the page in RAM.

If the page is not in RAM, then it's a page fault and program look for the data in secondary storage.

So, typical flow would be

Page Requested >> TLB miss >> cache miss >>main memory>> page fault >> looks in secondary memory.

Here,

Main memory access time= 30 ns

Page fault rate=.01%

page fault service time= 12ns

TLB access time=7 ns

TLB hit rate= .95%

TLB miss rate =1-.95=.05%

cache access time = 15 ns

cache miss rate= .3%

cache hit rate = 1-.3=.97%

So,

a) TLB hit time= TLB access time = 7 ns

cache hit time = TLB hit rate * TLB access time + TLB miss rate * ( TLB access time + cache hit time)

= .95 * 7 + .05 * (7+15)

= 7.75 ns

b) EAT for TLB hit= 7ns

Total EAT = TLB hit rate *( TLB access time + Cache hit rate * cache access time + cache miss rate * (cache + main memory access time))+ TLB miss rate ( TLB access time + main memory access time + cache hit rate * cache access time + cache miss rate ( cache + main memory access time))

= .95 *( 7 + (.97*15) + .03(15+30))+ .05*(7+30+(.97*15) + .03 ( 15 + 30))=24.4 ns

8 0
3 years ago
1. Springs____________<br> energy when compressed<br> And _________energy when they rebound.
densk [106]

Answer:

Springs store energy when compressed and release energy when they rebound

Explanation:

6 0
3 years ago
Which situation might cause potential hazards at a construction site?
nata0808 [166]

Answer:

D. protruding steel rebars .. #answerwithquality #BAL

5 0
3 years ago
Read 2 more answers
Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity
IceJOKER [234]

Explanation:

<u>(a)</u>

<u>The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity.</u> For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat.<u> The SI unit of heat capacity is W/m.K.</u>

The expression for thermal conductivity is:

q=-\kappa \bigtriangledown T

Where,

q is the heat flux

\kappa is the thermal conductivity

\bigtriangledown T is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

C=\frac{E}{\Delta T}

Where,

C is the Heat capacity

E is the energy absorbed/released

\Delta T is the change in temperature

<u>(c)</u>

<u>Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.</u>

The expression for thermal diffusivity is:

\alpha=\frac{\kappa}{C_p \times \rho}

Where,

\alpha is thermal diffusivity

\kappa is the thermal conductivity

C_p is specific heat capacity at constant pressure

\rho is density

6 0
3 years ago
Which branch of engineering studies the physical behavior of metallic elements?
olga55 [171]

Material engineering studies the physical behavior of metallic elements.

Answer: Option C

<u>Explanation: </u>

Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.

Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.

6 0
3 years ago
Read 2 more answers
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