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3 years ago

When you heat a pot on a stove, the handle gets warm. which type of heat transfer is possible

Computers and Technology

2 answers:

valina [46]3 years ago

5 0

Answer:

conduction

Explanation:

The handle will warm up until it's total heat losses equal the total heat coming in. Heat comes in mostly by conduction from the body of the pot.

bulgar [2K]3 years ago

4 0

Answer:

Conduction, I think

Explanation:

Conduction is the heat transfer through objects touching.

Hope this helps :)

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3 years ago

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4 years ago

In 2000, operating a basic internet application cost businesses approximately _____ per month. In mid-2016, operating the same a

anyanavicka [17]

Answer:

In 2000, a basic internet application cost businesses approximately $150,000 per month. In mid-2016, operating the same application in Amazon's cloud cost approximately $1,000 per month.

Explanation:

The role of cloud services cannot be overemphasized. Providing Infrastructure as a service (IaaS), Platform as a service (PaaS) and Software as a service (SaaS) cost a lot as there were very few cloud service providers and most services were yet to be integrated to the cloud.

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3 years ago

¿Cuáles son los 5 Procesadores de textos on line más usados?

labwork [276]

Answer:

5 Procesadores de textos

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3 years ago

Suppose the daytime processing load consists of 65% CPU activity and 35% disk activity. Your customers are complaining that the

bekas [8.4K]

Answer:

The answer to this question can be described as follows:

Explanation:

Given data:

Performance of the CPU:

The Fastest Factor Fraction of Work:

$f_1=65 \% \\\\=\frac{65}{100} \\\\ =0.65$

Current Feature Speedup:

$K_1=$ 1.5

CPU upgrade=6000

Disk activity:

The quickest part is the proportion of the work performed:

Current Feature Speedup:

$k_2=3$

Disk upgrade=8000

System speedup formula:

$s=\frac{1}{(1-f)+(\frac{f}{k})}$

Finding the CPU activity and disk activity by above formula:

CPU activity:

$S_{CPU}=\frac{1}{(1-f_1)+(\frac{f_1}{k_1})} \\\\=\frac{1}{(1-0.65)+(\frac{0.65}{1.5})} \\\\=1.276 \% ...\rightarrow (1) \\$

Disk activity:

$S_{DISK} = (\frac{1}{(1-f_2)+\frac{f_2}{k_2}}) \\\\ S_{DISK} = (\frac{1}{(1-0.35)+\frac{0.35}{3}}) \\\\ = -0.5\% .... \rightarrow (2)$

CPU:

Formula for CPU upgrade:

$= \frac{CPU \ upgrade}{S_{CPU}}\\\\= \frac{\$ 6,000}{1.276}\\\\= 4702.19....(3)$

DISK:

Formula for DISK upgrade:

$=\frac{Disk upgrade} {S_{DISK}}\\\\= \frac{\$ 8000}{-0.5 \% }\\\\= - 16000....(4)$

equation (3) and (4),

Thus, for the least money the CPU alternative is the best performance upgrade.

From (3) and (4) result,

The disc choice is therefore the best choice for a quicker system if you ever don't care about the cost.

The break-event point for the upgrades:

=4702.19 x-0.5

= -2351.095

From (2) and (3))

Therefore, when you pay the sum for disc upgrades, all is equal $ -2351.095

4 0

3 years ago