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3 years ago

explain how an earth wire in an appliance can help to prevent damage or injury to users when a fault occurs in the appliance.

Physics

1 answer:

Umnica [9.8K]3 years ago

3 0

Answer:

Simply put: the grounding wire is a way of electrical protection. Its function is to quickly connect the current to the earth through the grounding wire when your electrical equipment is leaking or inductively charged, so that the equipment shell is no longer charged, thereby ensuring the safety of the equipment behind the personnel.

Explanation:

For example, due to poor insulation performance of household appliances or humid environment, the shell will be charged with static electricity, and electric shock accidents may occur in serious cases. In order to avoid accidents, a wire can be connected to the metal shell of the electrical appliance, and the other end of the wire can be connected to the ground. Once the electrical leakage occurs, the grounding wire will bring static electricity into the ground and release it. In addition, it is more important for electrical maintenance personnel to use electric soldering irons to solder circuits, sometimes because the electric soldering irons are charged and break down and damage the integrated circuits in the electrical appliances. Friends who use computers sometimes ignore the grounding of the mainframe. In fact, connecting a ground wire to the mainframe of the computer can prevent the occurrence of crashes to a certain extent.

i hope it help you out ο(=•ω＜=)ρ⌒☆

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A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$

Charge of second Particle $q_2=-8\ \mu C$

distance between them $d=4\ cm$

$k=9\times 10^{9}$

magnetic field due to first charge at mid-way between two charged particles is

$E_1=\frac{kq_1}{r^2}$

$r=\frac{d}{2}=\frac{4}{2}=2\ cm$

$E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_1=27\times 10^7\ N/C$ (away from it)

Electric field due to $q_2=-8\ \mu C$

$E_2=\frac{kq_2}{r^2}$

$E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_2=-18\times 10^7\ N/C$ (towards it)

$E_{net}=E_1+E_2$

$E_{net}=9\times 10^7\ N/C$ (away from first charge)

8 0

3 years ago

Help !!!! Estimate the number of breaths taken by a person during 44 years?

Lisa [10]

44 x 12. I got the 12 from the total of 12 months in a year.

44 > 40

x
12 > 10
----------
The way my teacher taught me how to estimate is look at the neighbor to 44 and 12. The only time 44 can become 50, is when the neighbor is 5 or up. Same thing for 12. Now, multiply 40 and 10.

40 x 10 = 400.

Therefore, your estimate is 400.

The real answer is 520 breaths.

6 0

3 years ago

Two loudspeakers emit 600 Hz Hz notes. One speaker sits on the ground. The other speaker is in the back of a pickup truck. You h

Firdavs [7]

Answer:

The truck's speed is 4.04 m/s.

Explanation:

Given that,

Emit frequency = 600 Hz

Beat = 7.00 beat/sec

We need to calculate the truck's speed

Using formula of speed

$\text{frequency observed}=\text{frequency emitted}\times\dfrac{v}{v+v_{source}}$

Where, v = speed of sound

Put the value into the formula

$(600-7)=600\times(\dfrac{343}{343-v_{truck}})$

$v_{truck}=\dfrac{600\times343-593\times343}{593}$

$v_{truck}=4.04\ m/s$

Hence, The truck's speed is 4.04 m/s.

3 0

3 years ago

what is the electric potential at point A in the electric field created by a point charge of 5.5 • 10^-12 C? estimate k as 9.00

Hitman42 [59]

The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v

<u>Explanation</u>:

Given data,

charge = 5.5 x 10¹² C

k =9.00 x 10⁹

The electric potential V of a point charge can found by,

V= kQ / r

Assuming, r=5.00×10⁻² m

V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m

V= 49.5 x 10⁻³/ 5.00×10⁻²

Electric potential V= 0.099 x 10⁻¹v

3 0

3 years ago

A bus starts from rest.if the acceleration is 2m/s square, find

MrMuchimi

Answer:

The velocity after 2 seconds can be found through:

V = u +a*t

Where V is final velocity, u is initial velocity, a is acceleration and t is time.

V = 0 + 2* 2= 4 meters/second

The distance (s) can be found through:

V^2= u^2 +2*a* s

Where V is final velocity, u is initial velocity, a is acceleration.

4^2= 0^2 + 2 *2*s

16= 0 + 4s

s= 4 meters

Distance (s) can also be found through:

s= ut + 1/2 at^2

s= 0+ 1/2 *2*2^2= 1 *2*2

s= 4 meters

Explanation:

3 0

2 years ago