All categories

3 years ago

Explain nanotechnology. Its advantages, disadvantages with examples or application?

Physics

1 answer:

Volgvan3 years ago

7 0

Answer:

PURPOSE

This information bulletin clarifies the Department of Building and Safety’s (LADBS) requirement

relating to the equipment (product) and wiring approval as delineated in Sections 110.2, 110.3,

93.0401, 93.0402, and 93.0403 of the 2017 edition of the City of Los Angeles Electrical Code (LAEC).

DEFINITIONS

Approved - Acceptable to authority having jurisdiction. {California Electrical Code (CEC), Article 100}

Authority Having Jurisdiction – An organization, office, or individual responsible (i.e., LADBS

Electrical Testing Laboratory or Inspection) for enforcing the requirements of a code or standard, or

for approving equipment, material, an installation, or a procedure. (CEC, Article 100)

Equipment - A general term, including material, fittings, devices, appliances, luminaires, apparatus,

machinery, and the like used as part of, or in connection with, an electrical installation. (CEC, Article

100)

Identified - Recognizable as suitable for the specific purpose, function, use, environment, application,

and so forth, where described in particular code requirement. (CEC, Article 100)

Labeled - Equipment or material to which has been attached a label, symbol, or other accepted

identifying mark (i.e., embossed laboratory logo) of a recognized (approved) testing agency (see

definition) and concerned with product evaluation, that maintains periodic inspection of production of

labeled equipment or materials, and by whose labeling the manufacturer indicates compliance with

recognized national safety standards (see definition). (CEC, Article 100)

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A centrifuge rotor rotating at 10,000 rpm is shut off and is eventually brought to rest by a frictional force of 1.20m n. if the

Pani-rosa [81]

<span>Answer: The moments of inertia are listed on p. 223, and a uniform cylinder through its center is: I = 1/2mr2 so I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2 Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration: t = Ia -1.20 Nm = (0.0120984 kgm2)a a = -99.19 rad/s/s Now we have a kinematics question to solve: wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s w = 0 a = -99.19 rad/s/s Let's find the time first: w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s t = 10.558 s = 10.6 s And the displacement (Angular) Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form s = (u+v)t/2 Which is q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians But the problem wanted revolutions, so let's change the units: q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>

6 0

3 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

Alina [70]

Thank you for posting your question here at brainly. But your question seems incomplete. I will assume you based the situation below:

<span>An electrons moves at 2.0x10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.4x10^-2 T.

The </span> largest possible magnitude of the acceleration of the electron due to the magnetic field is <span>= 2.6 × 10 ¹⁶ m / s ²</span>

6 0

2 years ago

What is the value of x in the equation below 1+2e^x+1=9

GuDViN [60]

<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of $x$ :