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3 years ago

15

Opposition to current flow, restricts or resists current flow

1 answer:

BlackZzzverrR [31]3 years ago

8 0

Answer:

The answer is c-resistance

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_______ is a material property that pertains to local resistance to plastic deformation, such as scratching or denting. It is of

Readme [11.4K]

Answer: hardness

Explanation:

Hardness is a measure of a material's ability to resist plastic deformation. In other words, it is a measure of how resistant material is to denting or scratching. Diamond, for example, is a very hard material. It is extremely difficult to dent or scratch a diamond. In contrast, it is very easy to scratch or dent most plastics.

7 0

3 years ago

A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?

Illusion [34]

Answer:

0.0659 A

Explanation:

Given that :

$I_{0} = 6nA$ ( saturation current )

at 25°c = 300 k ( room temperature )

n = 2 for silicon diode

Determine the saturation current at 100 degrees = 373 k

Diode equation at room temperature = I = Io $\frac{V}{e^{0.025*n} }$

next we have to determine the value of V at 373 k

q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1

Given that I is constant

Io = $\frac{e^{0.025*2} }{31.08}$ = 0.0659 A

3 0

3 years ago

A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b

Marysya12 [62]

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= $\frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}$

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= $\frac{8640*100-8640*20}{20*10^5}$

k= 0.35 per day

6 0

3 years ago

I WILL GIVE BRAINLIEST IF ANSWER FAST What is the measurement on this Dial Caliper?

garik1379 [7]

Answer:

b i think i dont see any dial caliper

Explanation:

8 0

3 years ago

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

2 years ago