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3 years ago

.) If the charges attracting each other in the problem above have equal magnitude, what is the magnitude of each charge?

Engineering

1 answer:

Sedaia [141]3 years ago

8 0

Answer:

Not seeing any other information, the best answer I can give is 2m.

Explanation:

M = magnitude

You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.

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Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizon

Ulleksa [173]

The complete Question is:

Airflow through a long, 0.15-m-square air conditioning duct maintains the outer duct surface temperature at 10°C. If the horizontal duct is uninsulated and exposed to air at 35°C in the crawlspace beneath a home, what is the heat gain per unit length of the duct? Evaluate the properties of air at 300 K. For the sides of the duct, use the more accurate Churchill and Chu correlations for laminar flow on vertical plates.

What is the Rayleigh number for free convection on the outer sides of the duct?

What is the free convection heat transfer coefficient on the outer sides of the duct, in W/m2·K?

What is the Rayleigh number for free convection on the top of the duct?

What is the free convection heat transfer coefficient on the top of the duct, in W/m2·K?

What is the free convection heat transfer coefficient on the bottom of the duct, in W/m2·K?

What is the total heat gain to the duct per unit length, in W/m?

Answers:

- 7709251 or 7.709 ×10⁶

- 4.87

- 965073

- 5.931 W/m² K

- 2.868 W/m² K

- 69.498 W/m

Explanation:

Find the given attachments for complete explanation

4 0

3 years ago

Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate

Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy + flow work

$E = \dot m ( \frac{v^2}{2] + zg + p\nu)$

$E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})$

$57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})$

solving for flow rate

$\dot m = 99.977we know that [tex]\dot m = \rho AV$

$\dot m = 780 \frac{\pi}{4} D^2\times 16$

solving for d

$99.97 = 780 \times \frac{\pi}{4} D^2\times 16$

d = 0.090 m

so radius = 0.045 m

3 0

3 years ago

In Example 2-1, part c, the data were represented by the normal distribution function f(x)=0.178 exp(-0.100(x-451)2 Use this dis

valkas [14]

Answer:

P ( 2.5 < X < 7.5 ) = 0.7251

Explanation:

Given:

- The pmf for normal distribution for random variable x is given:

f(x)=0.178 exp(-0.100(x-4.51)^2)

Find:

the fraction of individuals demonstrating a response in the range of 2.5 to 7.5.

Solution:

- The random variable X follows a normal distribution with mean u = 4.51, and standard deviation s.d as follows:

s.d = sqrt ( 1 / 0.1*2)

s.d = sqrt(5) =2.236067

- Hence, the normal distribution is as follows:

X ~ N(4.51 , 2.236)

- Compute the Z-score values of the end points 2.5 and 7.5:

P ( (2.5 - 4.51) / 2.236 < Z < (7.5 - 4.51 ) / 2.236 )

P ( -0.898899327 < Z < 1.337168651 )

- Use the Z-Table for the probability required:

P ( 2.5 < X < 7.5 ) = P ( -0.898899327 < Z < 1.337168651 ) = 0.7251

6 0

3 years ago

The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat

lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar

Saturation temperature corresponding to saturation pressure =40°C

$h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}$

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

$Total\ heat\ =C_p\Delta T+\Delta h$

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

$h_1=\dfrac{\Delta h}{\Delta T}$

$h_1=\dfrac{2406}{20}$

$h_1=120.3\frac{KJ}{kg-m^2K}$

Now film coefficient after inclusion of sensible heat

$h_2=\dfrac{total\ heat}{\Delta T}$

$h_2=\dfrac{2,544}{20}$

$h_2=127.2\frac{KJ}{kg-m^2K}$

$So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100$

$=\dfrac{127.2-120.3}{120.3}\times 100$

=5.75 %

So Percentage change 5.75 %.

3 0

3 years ago

In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise

lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

$\dfrac{N_{sun}-N}{N_{ring}-N}=-R$

$\dfrac{5-N}{12-N}=-8$

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0

3 years ago