Ask question

Ask question

All categories

1 year ago

10

A microscope illuminator uses a transformer to step down the 120 V AC of the wall outlet to power a 12.0 V,50 W microscope bulb.

What is the resistance of the bulb filament

1 answer:

Anna35 [415]1 year ago

5 0

Answer:2.88 ohms

Explanation:

R= V^2 / P

12^2/50

144/50

2.88 ohms

You might be interested in

Three-dimensional measuring references all of these EXCEPT:

cricket20 [7]

Except the Table of Contents

4 0

2 years ago

Is the flow of power reversible in a worm and wheel gear

nikitadnepr [17]

The flow of power cannot be reversed since the slider could not move the worm gears. Since the input has one continuous tooth and the output has not teeth there is no gear ratio and no change in torque and speed.

8 0

2 years ago

Can you list three ways that real life earthquake conditions may differ from those made by a shaking table

jolli1 [7]

Answer:

No

Explanation:

8 0

3 years ago

A dryer is shaped like a long semi-cylindrical duct of diameter 1.5 m. The base of the dryer is occupied with water-soaked mater

Anestetic [448]

Answer:

0.0371 kg/s.m

Explanation:

From the given information, let's have an imaginative view of the semi-cylinder; (The image is shown below)

Assuming the base surface of both ends of the cylinder is denoted by:

$A_1 \ and \ A_2$

Thus, using the summation rule, the view factor $F_{11$ and $F_{12$ is as follows:

$F_{11}+F_{12}=1$

Let assume the surface (1) is flat, the $F_{11} = 0$

Now:

$0+F_{12}=1$

$F_{12}=1$

However, using the reciprocity rule to determine the view factor from the dome-shaped cylinder $A_2$ to the flat base surface $A_1$ ; we have:

$A_2F_{21} = A_{1}F_{12} \\ \\ F_{21} = \dfrac{A_1}{A_2}F_{12}$

Suppose, we replace DL for $A_1$ and

$A_2$ = $\dfrac{\pi D}{2}$

Then:

$F_{21} = \dfrac{DL}{(\dfrac{\pi D}{2}) L} \times 1 \\ \\ =\dfrac{2}{\pi} \\ \\ =0.64$

Now, we need to employ the use of energy balance formula to the dryer.

i.e.

$Q_{21} = Q_{evaporation}$

But, before that; let's find the radian heat exchange occurring among the dome and the flat base surface:

$Q_{21}= F_{21} A_2 \sigma (T_2^4-T_1^4) \\ \\ Q_{21} = F_{21} \times \dfrac{\pi D}{2} \sigma (T_2^4 -T_1^4)$

where;

$\sigma = Stefan \ Boltzmann's \ constant$

$T_1 = base \ temperature$

$T_2 = temperature \ of \ the \ dome$

∴

$Q_{21} = 0.64 \times (\dfrac{\pi}{2}\times 1.5) \times 5.67 \times 10^4 \times (1000^4 -370^4)\\ \\ Q_{21} = 83899.15 \ W/m$

Recall the energy balance formula;

$Q_{21} = Q_{evaporation}$

where;

$Q_{evaporation} = mh_{fg}$

here;

$h_{fg}$ = enthalpy of vaporization

m = the water mass flow rate

∴

$83899.15 = m \times 2257 \times 10^3 \\ \\ m = \dfrac{83899.15}{ 2257 \times 10^3 }\\ \\ \mathbf{m = 0.0371 \ kg/s.m}$

6 0

2 years ago

) An ammonia nitrogen analysis performed on a wastewater sample yielded 30 mg/L as nitrogen. If the pH of the sample was 8.5, de

Delicious77 [7]

Answer:

$NH_4^+ = 2.5 mg/lt$

Explanation:

Given data:

Ammonia Nitrogen 30 mg/L

pH = 8.5

-log[H +] = 8.5

[H +] = 10^{-8.5}

$NH_4 ^{+} ⇄ H^{+} + NH_3$

Rate constant is given as

$K_a = \frac{[H^{+}] [NH_3]}{NH_4^{+}}$ ...........1

$K_a = 5.6 \times 10^{-10}$

Total ammonia as NItrogen is given as 30 mg/l

$\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}$

= $\frac{100}{\frac{NH_4^+}{NH_4^+} +\frac{NH_3^+}{NH_4^+}}$

$= \frac{100}{1+ \frac{NH_3^+}{NH_4^+}}$ .....2

from equation 1 we have

$\frac{NH_3^+}{NH_4^+} =\frac{K_a}{[H^+]} = \frac{5.6\times 10^{-10}}$ {10^{8.5}}

plug this value in equation 2 we get

$\%NH_4^{+} = 84.96 \%$

Total ammonia as N = 30 mg/lt

$NH_4^+ = \frac{84.96}{100} \times 30 = 25.5 mg/lt$

7 0

2 years ago