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ss7ja [257]
3 years ago
9

Importance of electrical symbol​

Physics
1 answer:
Vsevolod [243]3 years ago
8 0
The electrical symbols are very important especially when fixing electrical appliances because it tells you where the wire of neutral and live go.
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What keeps the balanced rock balanced
pogonyaev
Hello,

Here is your answer: 

The proper answer to this question is "because of there substantial size the rock rests on another rock which keeps it balanced".

If you need anymore help feel free to ask me!

Hope this helps!
5 0
3 years ago
How much heat is needed to boil 120 kg of water ?
nekit [7.7K]
Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)

Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ

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7 0
2 years ago
Help with this I can’t come up with anything
kakasveta [241]
Let’s say you have a spring. You press on the spring with your finger. The spring goes down. This is the action force. Then, the spring goes back up after you take your finger off of it. This is known as the reaction force.
3 0
3 years ago
Disadvantage of entrepreneurship​
pochemuha

Answer:

limited liability, limitation in expansion, risk bearing, problem of continuity,

7 0
3 years ago
Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q /  (5 x 10⁻² )²

E₁  = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁  = KQ x 10⁴  i / 25  

Similarly electric field due to charge 3Q near 2Q

=  3KQ x 10⁴  i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴  j / 25

Similarly electric field due to charge 4Q near 2Q

=  4KQ x 10⁴  j / (25 x 2 )

= 2 KQ x 10⁴  / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴  ( i + j )  / 25 x √2

Total field =  KQ x 10⁴  i / 25 + 3KQ x 10⁴  j / 25 + 2 KQ x 10⁴  ( i + j )  / ( 25 x √2 )

= KQ x 10⁴  [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j )  N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j )  x 2Q  N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0
3 years ago
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