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3 years ago

15

A disk of radius 2.1 cm has a surface charge density of 5.6 µC/m2 on its upper face. What is the magnitude of the electric field

produced by the disk at a point on its central axis at distance z = 9.5 cm from the disk?

1 answer:

Assoli18 [71]3 years ago

8 0

Answer:

$=6.3*10^3 N/C$

Explanation:

solution:

from this below equation (1)

$E=$ σ/2εo $(1-\frac{z}{\sqrt{z^2-R^2} } )$ ...........(1)

we obtain:

$=5.6*10^-6 \frac{c}{m^2} /2(8.85*10^-12\frac{c^2}{N.m^2} ).(1-\frac{9.5 cm}{\sqrt{9.5^2-2.1^2} } )$

$=6.3*10^3 N/C$

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I = 48 mA, R = 1125 2. What is Vs ?

german

54 volts

Ohms law. E= I x R

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3 years ago

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough

jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection

Explanation:

We are given:

Plane strain fracture toughness K $= 98.9 MPa \sqrt{m}$

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

$oc = \frac{K}{Y \sqrt{pi * a}}$

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

$a= \frac{1}{pi} * [\frac{k}{y*a}]^2$

Substituting figures in the expression above, we have:

$= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2$

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection

3 0

3 years ago

Read 2 more answers

The time factor for a doubly drained clay layer

Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

$T_v=\frac{\pi }{4}(\frac{U}{100})^2$

Solving for 'U' we get

$\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%$

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i) $\frac{z}{H}=0.25=U=0.71$ = 71% consolidation

ii) $\frac{z}{H}=0.5=U=0.45$ = 45% consolidation

iii) $\frac{z}{H}=0.75U=0.3$ = 30% consolidation

Part b)

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm$

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

$T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59$

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm$

5 0

3 years ago

I wish to have a computer whose machine-level instructions are all 32 bits each. If I want to have all instructions of the form

melisa1 [442]

Answer:

Maximum number that can be represented by 13 bits is 8192 Instructions

Explanation:

number of instructions = 1000

number of bits = log(1000) x number of register

= 6 bits

Since the complete instruction must have 32 bits, then

remaining number of bits = 32 - 6 = 236

number of registers in instruction = 2

number of bits per register = 26/2 = 13

Maximum number that can be represented by 13 bits = $2^{n}$

= 2¹³ = 8192

4 0

3 years ago

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

Anettt [7]

Answer:

$Q=4.98\times 10^{-3}\ m^3/s$

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

For Copper tube is 3/4 standard type K drawn tube

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

$\mu =0.00164\ Pa.s$

Pressure difference given as

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

Where

L is length of tube

μ is dynamic viscosity

Q is volume flow rate

d is inner diameter of tube

ΔP is pressure drop

Now by putting the values

$\Delta P=\dfrac{128\mu QL}{\pi d_i^4}$

$130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}$

$Q=4.98\times 10^{-3}\ m^3/s$

So flow rate is $Q=4.98\times 10^{-3}\ m^3/s$

7 0

3 years ago