All categories

3 years ago

What are supercapacitors ?

Engineering

2 answers:

OLEGan [10]3 years ago

6 0

Answer:

A supercapacitor, also called an ultracapacitor, is a high-capacity capacitor with a capacitance value much higher than other capacitors, but with lower voltage limits, that bridges the gap between electrolytic capacitors and rechargeable batteries.

Explanation:

Masja [62]3 years ago

6 0

Answer:A supercapacitor, also called an ultracapacitor, is a high-capacity capacitor with a capacitance value much higher than other capacitors, but with lower voltage limits, that bridges the gap between electrolytic capacitors and rechargeable batteries

Explanation:

You might be interested in

Showing all of your work and algebra,generate an approximate expression for T as a function ofthe other variables. (b) Explain w

shusha [124]

Answer:

Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.

5 0

3 years ago

What do you guys like in engineering

Drupady [299]

Answer:

building lol and actually workin

Explanation:

3 0

3 years ago

Read 2 more answers

A construction crew lifts approximately 400 lb. of material several times during a day from a flatbed truck to a 25 ft. rooftop.

Irina18 [472]

Answer:

2ib

Explanation:

if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8

there goes your answer.

5 0

2 years ago

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

$F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}$

Substituting figures in the equation we have:

$= \frac{1}{1+0.1(2.5-1)+0(2-1)}$

= 0.75

Let's now calculate equivalent flow rate of the car using:

$Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}$

$= \frac{7500}{0.9*4*0.75*1}$

= 2777.7 pc/h/en

Calculating for traffic density, we have:

$D = \frac{Vp}{FFS}$

$D = \frac{2777.7}{70}$

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0

3 years ago

With a very precise volumetric measuring device, the volume of a liquid sample is determined to be 6.321 L (liters). Three stude

zheka24 [161]

Answer:

See explanation

Explanation:

Solution:-

- Three students measure the volume of a liquid sample which is 6.321 L.

- Each student measured the liquid sample 4 times. The data is provided for each measurement taken by each student as follows:

Students

Trial A B C

1 6.35 6.31 6.38

2 6.32 6.31 6.32

3 6.33 6.32 6.36

4 6.36 6.35 6.36

- We will define the two terms stated in the question " precision " and "accuracy"

- Precision refers to how close the values are to the sample mean. The dense cluster of data is termed to be more precise. We will use the knowledge of statistics and determine the sample standard deviation for each student.

- The mean measurement taken by each student would be as follows:

$E ( A ) = \frac{6.35 +6.32+6.33+6.36}{4} \\\\E ( A ) = 6.34\\\\E ( B ) = \frac{6.31 +6.31+6.32+6.35}{4} \\\\E ( B ) = 6.3225\\\\E ( C ) = \frac{6.38 +6.32+6.36+6.36}{4} \\\\E ( C ) = 6.355\\$

- The precision can be quantize in terms of variance or standard deviation of data. Therefore, we will calculate the variance of each data:

$Var ( A ) = \frac{6.35^2+6.32^2+6.33^2+6.36^2}{4} - 6.34^2\\\\Var ( A ) = 0.00025\\\\Var ( B ) = \frac{6.31^2+6.31^2+6.32^2+6.35^2}{4} - 6.3225^2\\\\Var ( B ) = 0.00026875\\\\Var ( C ) = \frac{6.38^2+6.32^2+6.36^2+6.36^2}{4} - 6.355^2\\\\Var ( C ) = 0.000475\\$

- We will rank each student sample data in term sof precision by using the values of variance. The smallest spread or variance corresponds to highest precision. So we have:

Var ( A ) < Var ( B ) < Var ( C )

most precise Least precise

- Accuracy refers to how close the sample mean is to the actual data value. Where the actual volume of the liquid specimen was given to be 6.321 L. We will evaluate the percentage difference of sample values obtained by each student .

$P ( A ) = \frac{6.34-6.321}{6.321}*100= 0.30058\\\\P ( B ) = \frac{6.3225-6.321}{6.321}*100= 0.02373\\\\P ( C ) = \frac{6.355-6.321}{6.321}*100= 0.53788\\$

- Now we will rank the sample means values obtained by each student relative to the actual value of the volume of liquid specimen with the help of percentage difference calculated above. The least percentage difference corresponds to the highest accuracy as follows:

P ( B ) < P ( A ) < P ( C )

most accurate least accurate

7 0

3 years ago