Answer:
The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²
Explanation:
The electric field intensity due to a thin conducting sheet is given by the following formula:
Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)
From this formula:
Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)
We have the following data:
Electric Field Intensity = 1.5 N/C
Permittivity of free space = 8.85 x 10^-12 C²/N.m²
Therefore,
Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)
<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>
Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².
Answer:
Architect Registration Examination (ARE)
Explanation:
In this scenario, Joseph wants to practice architecture. A compulsory assessment administered by National Council of Architectural Registration Boards (NCARB) which he has to complete is an Architect Registration Examination.
An Architect Registration Examination (ARE) refers to the professional licensure examination to be taken by anyone who intends to practice architecture in the United States of America, Puerto Rico, Guam, Canada, US Virgin Islands.
The main purpose of the Architect Registration Examination is to assess an architect's knowledge, skills, and abilities on architectural best practices, procedures and services. Also, it focuses on areas relating to a building's safety, soundness and health impact on the habitants. Therefore, ARE is a prerequisite for practicing architecture across the United States of America jurisdiction.
Once an architect passes the examination, he or she would be given a license to practice architecture.
<em>Hence, for Joseph to practice architecture he must take the Architect Registration Examination developed and administered by National Council of Architectural Registration Boards (NCARB). </em>
Answer:
There were a lot of great engineering achievements presented in the 20th century. To name some, we have the electricity, airplane, radio and television, water supply and distribution, computers, television, X-ray imaging, nuclear technologies, and of course the Internet.
Answer:
The time necessary to purge 95% of the NaOH is 0.38 h
Explanation:
Given:
vfpure water(i) = 3 m³/h
vNaOH = 4 m³
xNaOH = 0.2
vfpure water(f) = 2 m³/h
pwater = 1000 kg/m³
pNaOH = 1220 kg/m³
The mass flow rate of the water is = 3 * 1000 = 3000 kg/h
The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg
When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg
The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³
The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h
The initial void ratio is the <em>parameter </em>which is used to show the structural foundations for each <em>specimen of sand </em>so that the method and speed of compression would be <em>measured</em>.
Relative density is the mass per unit volume of each specimen of sand which is <em>measured </em>and it has to do with the<em> relative ratio</em> of the density of the sand.
Unit weight is the the exact weight per cubic foot of the sand which is measured.
Please note that your question is incomplete so I gave you a general overview to help you better understand the concept
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