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3 years ago

11

To estimate the distance in kilometers of a flash of lightning, count the number of seconds between seeing the flash and hearing

the accompanying thunder, then divide by:__________

1 answer:

Triss [41]3 years ago

8 0

Answer:

Divide by 3

Explanation:

In order to estimate the distance traveled by a lightening flash in kilometers, we follow these simple steps:

Make a count of the number of seconds in between the period a flash occur and the thunder accompanied by the lightening flash is heard.
Dive the total number of seconds by 3 to get the distance traveled by the flash. This is because in order to cover 1 km, it roughly takes 3 seconds.

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Saturated ethylene glycol at 1 atm is heated by a horizontal chromiumplated surface which has a diameter of 200 mm and is mainta

Paha777 [63]

Here is the full question.

Saturated ethylene glycol at 1 atm is heated by a chromium-plated surface which is circular in shape and has a diameter of 200-mm and is maintained at 480 K.

At 470 K the properties of the saturated liquid are mu = 0.38 * 10-3 N. s/m^2, Cp = 3280 J/kg. K and Pr = 8.7. The saturated vapour density is p= 1.66 kg/m^3. Take the liquid to surface constants to be Cnb = 0.010 and m=4.1.

Estimate the heating power requirement and the rate of evaporation

What fraction is the power requirement of the maximum power associated with the critical heat flux

Answer:

The heating power requirement = 559.2 W

The rate of evaporation = $6.89*10^{-4}kg/s$

The fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

Explanation:

From the thermodynamics tables; we deduced the value for enthalpy at the pressure 1 atm and $T_{sat}$ = 470 K for the saturated ethylene glycol.

Value for enthalpy of formation $h_{fg}$ = 812 kJ/kg

Density of saturated ethylene glycol $\rho___l$ = 1111 kg/m³

Surface tension $\sigma = 32.7*10^{-3}N/m$

The heat flux can be calculated by using the formula:

$q"s = \mu___l}}}h_{fg}{[\frac{g(\rho{__l}- \rho{__v} }{\sigma} ]^{1/2} [\frac{C_p*\delta T_c}{C_{sf}*h_{fg}P_r} ]^3$

$= [0.38*10^{-3}\frac{NS}{m^2} *812*10^3\frac{J}{kg} (\frac{9.81m/s^2*(1111-1.66)kg/m^3}{32.7*10^{-3}N/m} )^{1/2}*(\frac{3280J/kg.K(480-470)K}{0.01*812*10^3\frac{J}{kg}*(8.7)^1 } )]$

= 308.56 × 576.6 × 0.1

= 1.78 × 10⁴ W/m²

Now; to find the heating power requirement; we have:

$q_{boil} = q__s }*A S$

= $1.78*10^4 \frac{W}{m^2}*(\frac{\pi}{4}*(0.2m))^2$

Thus, the heating power requirement = 559.2 W

The rate of evaporation is given as:

$m= \frac{q_{boil}}{h_[fg}}$

= $\frac{559.2}{812*10^3}$

= $6.89*10^{-4}kg/s$

Thus, the rate of evaporation = $6.89*10^{-4}kg/s$

To determine to what fraction in the power requirement of the maximum power is associated with the critical total flux ; we needed to first calculate the critical heat flux.

So, the calculation for the critical heat is given as: $q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}$

= $q"max = 0.149*812810^3* 1.66[ \frac{32.7*10^{-3}*9.8 (1111- 1.66}{1.66^2} ]^{1/4}$

= 200840.08 × 3.37

= 6.77 × 10⁵ W/m²

Finally, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is as follows:

= $\frac{q''s}{q''max}$

= $\frac{1.78*10^4}{6.77*10^5}$

= 0.026

Thus, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

7 0

3 years ago

Starting from the front door of your ranch house, you walk 50.0m due east to your windmill, and then you turn around and slowly

juin [17]

Answer:

velocity = 0.3m/s

speed = 1.21 m/s

Explanation:

The total time it takes to get from the front door to the bench is

t = 27 + 39 = 66 seconds

The net displacement from the front door to the bench is the distance from the front door to the windmill subtracted by the distance from the windmill to the bench

s = 50 - 30 = 20 m

So the average velocity is net displacement divided by total time

v = s / t = 20 / 66 = 0.3 m/s

The total distance from the front door to the bench is the sum of distance from the front door to the windmill and the distance from the windmill to the bench

S = 50 + 30 = 80 m

So the average speed is total distance divided by total time

v = s / t = 80 / 66 = 1.21 m/s

8 0

3 years ago

If an astronaut can jump straight up to a height of 0.6 m on earth, how high could he jump on the moon?

Zinaida [17]

Answer:

Explanation:

Given

Person on earth can jump to a height $s=0.6\ m$

initial velocity is u

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

final velocity is zero

$s=\frac{u^2}{2g}$

$0.6=\frac{u^2}{2g}-----1$

On moon surface acceleration due to gravity is \frac{1}{6}[/tex] th of earth gravity

so height attained is given by

$h=\frac{u^2\times 6}{2\cdot g}-----2$

divide 1 and 2 we get

$h=6\times 0.6=3.6\ m$

4 0

3 years ago

What can speed up the rate of a Chemical reaction???

Sav [38]

Most likely the variables involved.

7 0

3 years ago

An object's distance from a converging lens is 3.78 times the focal length. a) Determine the location of the image. Express the

Romashka-Z-Leto [24]

Answer:

a) v=1.36f

b)m=.36

c) real

d) Inverted

Explanation:

a)

In this question we have given,

object distance from converging lens,u=-3.78f

focal length of converging lens,=f

we have to find location of image,v=?

and magnification,m=?

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}- \frac{1}{u}$ .............(1)

put values of f u in equation (1)

we got,

$\frac{1}{f} =\frac{1}{v}- \frac{1}{-3.78f}$

$\frac{1}{f}-\frac{1}{3.78f} =\frac{1}{v}$

$\frac{2.78}{3.78f} =\frac{1}{v}$

or

v=1.36f

b) Magnification, $m=\frac{v}{u} \\m=\frac{1.36f}{3.78f}\\m=.36$

c)

Here the object is located further away from the lens than the focal point therefore image is real image and inverted.

d) image is inverted

7 0

3 years ago