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agasfer [191]
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?
Engineering
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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You are hitting a nail with a hammer (mass of hammer =1.8lb) the initial velocity of the hammer is 50 mph (73.33 ft/sec). The ti
Archy [21]

Answer:

The nail exerts a force of 573.88 Pounds on the Hammer in positive j direction.

Explanation:

Since we know that the force is the rate at which the momentum of an object changes.

Mathematically \overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}

The momentum of any body is defines as \overrightarrow{p}=mass\times \overrightarrow{v}

In the above problem we see that the moumentum of the hammer is reduced to zero in 0.023 seconds thus the force on the hammer is calculated using the above relations as

\overrightarrow{F}=\frac{m(\overrightarrow{v_{f}}-\overrightarrow{v_{i}})}{\Delta t}

\overrightarrow{F}=\frac{m(0-(-73.33)}{0.23}=\frac{1.8\times 73.33}{0.23}=573.88Pounds

6 0
3 years ago
A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome
Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

3 0
3 years ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0
3 years ago
Drag the tiles to the boxes to form correct pairs. Identify the designations of the three employees in an automobile company fro
aniked [119]

Answer:

is the fare of our responsibility towards

7 0
3 years ago
Please help me bro come on
melomori [17]
Answer:

12

Explanation:

5 • 3 = 15
3 • 3 = 9
4 • 3 = 12
6 0
3 years ago
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