All of the following have the same units except: Group of answer choices resistance capacitive reactance. inductance. impedance.

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

Assoli18 [71]

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:

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

$=6.3*10^3 N/C$

8 0

3 years ago

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

Which option identifies the tool best to use in the following scenario?

Lena [83]

Answer:

an Allen wrench

Explanation:

it is hexagonal

3 0

3 years ago

The Reynolds number is a dimensionless group defined for a fluid flowing in a pipe as Re Durho/μ whereD is pipe diameter, u is f

Gala2k [10]

Answer:

the flow is turbulent

Explanation:

The Reynolds number is given by

Re=ρVD/μ

where

V=fluid speed=0.48ft/s=0.146m/s

D=diameter=2.067in=0.0525m

ρ=density=0.805g/cm^3=805Kg/m^3

μ=0.43Cp=4.3x10^-4Pas

Re=(805)(0.146)(0.0525)/4.3x10^-4=14349.59

Re>2100 the flow is turbulent

Note: if you do not want to use a calculator you can use the graphs to calculate the Reynolds number according to their properties

8 0

3 years ago

A plane wall of length L, constant thermal conductivity k and no thermal energy generation undergoes one-dimensional, steady-sta

KIM [24]

Answer:

Temperature distribution is $T(x)=\dfrac{q}{k}(L-x)+T$

Heat flux=q

Heat rate=q A

Explanation:

We know that for no heat generation and at steady state

$\dfrac{\partial^2 T}{\partial x^2}=0$

$\dfrac{\partial T}{\partial x}=a$

$T(x)=ax+b$

a and are the constant.

Given that heat flux=q

We know that heat flux given as

$q=-K\dfrac{dT}{dx}$

From above we can say that

$a= -\dfrac{q}{K}$

Alos given that when x= L temperature is T(L)=T

$T= -\dfrac{q}{K}L+b$

$b=T+\dfrac{q}{K}L$

So temperature T(x)

$T(x)=-\dfrac{q}{K}x+T+\dfrac{q}{K}L$

$T(x)=\dfrac{q}{k}(L-x)+T$

So temperature distribution is $T(x)=\dfrac{q}{k}(L-x)+T$

Heat flux=q

Heat rate=q A (where A is the cross sectional area of wall)

6 0

3 years ago