All categories

3 years ago

In a reversible process both the system and surrondings can be returned to their initial states. a)-True b)-False

Engineering

1 answer:

Nuetrik [128]3 years ago

6 0

Answer:

a)-True

Explanation:

When the direction of a process is reversed through infinitesimal changes to the system. The infinitesimal changes are done by the surroundings. After the process has been reversed the system and surroundings go back to their original state.

The changes take an infinite amount of time so, it is not possible practically.

You might be interested in

(1.24) Consumer Reports is doing an article comparing refrigerators in their next issue. Some of the characteristics to be inclu

kondaur [170]

Answer:

“height is a quantitative variable ”

Explanation:

According to the question asked, answer is “height is a quantitative variable ”

Height is a quantitative variable because it is related to the measurement and in measurement, when we measure something we deal with number (numerical data)

Numerical data is a type of quantitative data that is why we say “height is a quantitative variable”

There are some other possible questions in the given paragraph which I would like to mention here, are as following:

Which are the categorical variables in the given report?

Answer: Energy star complaints

Top, Bottom or side-by-side freezer

Which are the quantitative variables in the given report?

Answer: Estimated Energy Consumption in kilowatts

Width, depth, and height in inches

Capacity in Cubic Feet

What are the individuals in the report?

Answer: The brand name and model

8 0

2 years ago

Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and

Pavlova-9 [17]

Answer:

Part A:

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

Explanation:

FP Instructions=50*106=5300

INT Instructions=110*106=11660

L/S Instructions=80*106=8480

Branch Instructions=16*106=1696

Calculating Execution Time:

Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

Execution Time= $\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}$

Execution Time= $2.7136*10^{-5}\ s$

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300* New\ CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12$

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time= $\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s$

$1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8$

CPI reduced by $1-\frac{0.8}{4} = 0.80$ =80%

Part C:

$New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4$

New Execution Time= $\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}$

New Execution Time= $\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s$

Increase in speed= $1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%$

8 0

3 years ago

The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm

Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0

3 years ago

A garden hose fills a 2-gallon bucket in 5 seconds. The number of gallons, g (y), is proportional to the number of seconds, t (x

stepladder [879]

Answer:

0.4 gallons per second

Explanation:

A function shows the relationship between an independent variable and a dependent variable.

The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.

The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).

Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.

Constant of proportionality = g / t = 2 / 5 = 0.4 gallons per second

5 0

3 years ago

To reduce the negative impacts of stress on your mental

soldi70 [24.7K]

Answer:

A&C

Explanation:

breathing deeply is relaxing

talking with a friend can helping

4 0

2 years ago

Read 2 more answers