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ozzi
2 years ago
8

a) A total charge Q = 23.6 μC is deposited uniformly on the surface of a hollow sphere with radius R = 26.1 cm. Use ε0 = 8.85419

X 10−12 C2/Nm2. What is the magnitude of the electric field at the center of the sphere? b) What is the magnitude of the electric field at a distance R/2 from the center of the sphere? c) What is the magnitude of the electric field at a distance 52.2 cm from the center of the sphere?
Engineering
1 answer:
dusya [7]2 years ago
4 0

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

<u>E = 0 N/C</u>

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

<u>E = 0 N/C</u>

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

<u>E = 7.78 x10^5 N/C</u>

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Mandarinka [93]

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3 years ago
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Water is the working fluid in an ideal Rankine cycle. Superheatedvapor enters the turbine at 10MPa, 480°C, and the condenser pre
choli [55]

Answer:

Explanation:

Given that:

Superheated vapor enters the turbine at 10 MPa, 480°C,

From the tables of superheated steam tables; the following values are obtained

h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K

Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state

s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}

s_2 =0.51624 + x(7.82)

s_2 =0.51624 + 7.82x

From the values obtained;

s_1 =s_2= 6.52846 \ kJ/kg.K

Therefore;

6.52846 = 0.51624+7.82x

6.52846 - 0.51624 = 7.82 x

6.01222  = 7.82 x

x = 6.01222/7.82

x = 0.7688

The enthalpy for this process at state (s_2) can be determined as follows:

h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 )  \\ \\ h_2 =2010.4084   \ kJ/kg

The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:

n_T = \dfrac{h_1-h_{2a}}{h_1-h_2}

0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}

0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}

0.8 * {1311.6116}= {3322.02-h_{2a}

1049.28928=  {3322.02-h_{2a}

h_{2a}=   {3322.02- 1049.28928

h_{2a}=   2272.73072 kJ/kg

The work pump is calculated by applying the formula:

w_p = v_{f  \  6 kpa} (p_4-p_3)

w_p = 0.0010062 * (10000-6)

w_p = 0.0010062 *9994

w_p = 10.0559628 \  kJ/kg

However;

w_p = h_4 -h_3

From the process;

h_3 = h_{f(6 kpa)} = 150.15 \  kJ/kg

10.0559628 = h_4 - 150.15

10.0559628+  150.15 = h_4

160.2059628= h_4

h_4= 160.2059628 \  kJ/kg

The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:

n_P = \dfrac{h_4-h_{3}}{h_{4a}-h_3}

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2 years ago
A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit
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Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

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3 years ago
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d
alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number (Re) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

Re=\frac{\rho v D}{\mu}

where \rhois the density of the fluid, \mu is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm

8 0
3 years ago
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