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3 years ago

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All questions and answer on ptessure

1 answer:

Lilit [14]3 years ago

3 0

Answer:

hi click on my profile to help me and y does none talk on brainly

Explanation:

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In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75C as it moves at 0.2 m/s through a

steposvetlana [31]

Answer:

The answer is " $12682.267\ \ \frac{W}{m^2}$ "

Explanation:

Calculating the mass flow rate of fluid:

$m= \rho AV$

$= \rho \frac{\pi}{4} D^2\ V\\\\= 100 \times \frac{\pi}{4} \times (0.0127)^2\times 0.2\\\\=0.0253 \ \frac{kg}{s}\\\\$

Calculating the amount of heat transfer.

$q =m\timesC_p(T_{m,0}-T_{m,i})$

$=0.0253 \times 4000 (75-25)\\\\=0.0253 \times 4000(50)\\\\=0.0253 \times 200,000 \\\\= 5060 \ W$

Calculating the required value for heat flux:

$q"=\frac{q}{A_s}$

$=\frac{q}{\pi DL}\\\\= \frac{5060}{\pi \times 0.0127 \times 10}\\\\= 12682.267 \frac{W}{m^2}\\$

3 0

3 years ago

One liter is equal to 100 milliliters true or false

IRINA_888 [86]

Answer:

No, a liter is equal to 1000 milliliters

7 0

3 years ago

Read 2 more answers

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

Which of the following statements is NOT correct?

Flauer [41]

The answer is
D:if the

6 0

3 years ago

Vector A has a magnitude 5 m, vector B has a magnitude 7 m. Which of the following cannot be the magnitude of the vector A + B?

bulgar [2K]

I think it should be 7m

7 0

4 years ago