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Ksju [112]
2 years ago
13

We all depend on electricity. Most electricity is created by electromagnetic generators at large power plants and distributed th

rough an electric grid. The electric grid in the United States is interconnected and includes more than 400,000 miles of electric transmission lines. However, the grid has not been well maintained, and the number and severity of power outages has been rising over the past decade. Because the grid is interconnected, a severe power outage in one area can quickly spread to other areas.
How long could you survive without electricity? What parts of your life would be affected by loss of electricity? Should you prepare for an electricity outage, and if so, how would you prepare? What backup system could your family or community install to generate limited amounts of electricity during an outage? How does this system create an electric force field and generate electric current?
Physics
1 answer:
Anuta_ua [19.1K]2 years ago
8 0

Answer:

1) not so long (maybe an hour or two)

2) access to information through the internet will be most affected if my computer and mobile phone run out of battery power.

3) yes, one should prepare for power outage. This can be done by having a standby alternative source of power like the use of inverters that stores electrical energy in form of chemical energy, and small internal combustion engine powered electric generators.

4) solar panels can be used to draw power from incident sun rays, this power can be stored in an inverter for future use in case of a power outage.

5) energy from the sun is converted into direct current which is then supplied to an accumulator in the opposite direction to its flow of current. When the energy is needed, it can be used directly, or converted to an alternating current. This is achieved by connecting its terminal to the supply. Electric field is generated by flow of ions and electrons within the working chemical (e.g lithium).

Explanation:

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Answer:

W = M g = 150 kg * 9.81 m/s^2 = 1470 N

You were only given 3 significant figures in the question.

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2 years ago
A first aid provider is trained to do the following except:
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A.
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3 years ago
This is a science Question. Please help.
sattari [20]

Answer:

B

Explanation:

This is a physics question, know that force is equals to mass divided by acceleration (acc.), so if the same force is applied, say 10 Newton and the mass of A is 2 and the mass of B is 4, then the acceleration of A is 0.2 and that of B is 0.4 by equating, and this applies to all cases.

6 0
3 years ago
A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp
AlekseyPX

Answer:

(A) 2.4 N-m

(B) 0.035kgm^2

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque \tau =Fr, here F is force and r is distance

So \tau =20\times 0.12=2.4Nm

(B) Moment of inertia is equal to I=\frac{1}{2}mr^2

So I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2

Torque is equal to \tau =I\alpha

So angular acceleration \alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2

(C) As the disk starts from rest

So initial angular speed \omega _{0}=0rad/sec

Time t = 4.6 sec

From first equation of motion we know that \omega =\omega _0+\alpha t

So \omega =0+68.571\times  4.6=315.426rad/sec

(D) Kinetic energy is equal to KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J

(E) From second equation of motion

\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad

3 0
3 years ago
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti
deff fn [24]

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

where E=electric field strength

A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

E=0.012\ N/C

and Electric field  by a uniformly charged sheet is given by

E=\frac{\sigma }{2\epsilon_0}

where \sigma=charge density

=\frac{\sigma }{\epsilon_0}

\sigma =0.012\times 8.85\times 10^{-12}

\sigma =2.12\times 10^{-13}\ C/m^2    

5 0
3 years ago
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