The magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2. 1 g that has been charged to -3. 0 NC is 6.86 × 
N/C.
An electric field is the physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field of a system of charged particles.
It is given that,
Mass of sphere, m = 2.1 g =0.0021kg
Charge,q = ₋3nC = ₋ 3 ₓ 
To balance the weight of a plastic sphere, we must determine the magnitude of the electric field. So,
a = g
E = 
E = 6860000 N/C
E = 6.86 × N/C
Hence, the magnitude of the electric field that balances its weight is 6.86 × N/C .
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Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2
20N/4 = 5N
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Answer:
Here we need to make parallel connection of two 80 ohm resistors to achieve 40 ohm net resistance.
Explanation:
As we know that the resistances in series add up directly and here we are given with only the resistors of 80 Ω.
So when we connect two resistors of 80 ohm in parallel we get the resultant of 40 ohm.
Mathematically:
gives us the only combination of two resistors in parallel.
Answer:
36 N
Explanation:
If the object of mass, m = 8 kg is swung in a horizontal circle of radius, r = 2m = length of string with tangential velocity v = 3 m/s, the tension in the string is the centripetal force which is T = mv²/r
= 8 kg × (3 m/s)²/2 m
= 4 kg × 9 m/s²
= 36 N
Answer:
Hari didn't plan to go abroad
Explanation:
Abroad planned to go to hari.
