A steel rule can be used to check for

Random ____ fluctuations in the early universe were stretched by a period of intense inflation, resulting in permanent density f

fgiga [73]

Answer:

Quantum

Explanation:

Appearance of energy particles from any where as allowed by uncertainty principle.

7 0

3 years ago

Which website suffixes are usually the least credible? Check all that apply.

zmey [24]

.com hope this helps

7 0

3 years ago

Read 2 more answers

An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q=1.0 μC, is located at t

vfiekz [6]

Answer:

work done by electric field is 0.06 J

Explanation:

Given data:

Two point charge is $+ 1\mu C and -1 \mu C$

0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)

-1 charge positioned is (0 cm , -1 cm, 0.00 cm)

E = 3.0\times 10^6 N/C

From above information, the distance between given two charges d = 2 cm

then d = 0.02m

work needed is W = q E d

$W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02$

W = 0.06 J

Therefore work done by electric field is 0.06 J

8 0

3 years ago

Student A says hazardous waste can take the form of solid, liquid, or gas. Student B says hazardous waste can only take the form

lina2011 [118]

Answer:

Student A

Explanation:

hope this helps have a great day

4 0

3 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago