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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Answer:
Answer is 717 N . m
Refer below for the explanation.
Explanation:
As per the question,
38 mm diameter shaft,
Yield strength 250 mpa,
P 240kn.
Refer to the picture for complete explanation.
Answer:
C ) 1500 Joules
Explanation:
In the attached image, we can see how the box was moved 10 meters by the force of 150 N.
The definition of work is W = F * d
therefore
W= 150*10 = 1500 J
Answer:
The net force is 15 newtons
The direction is to the right
Explanation:
Hope this helps
Acceleration is the change of velocity, and velocity is the change of distance. The opposite of finding change, or differentiation, is integration.
Acceleration = 1.3 m/s²
Velocity: ∫ 1.3 dx = 1.3x + c m/s
Distance: ∫ 1.3x dx = 1.3x²/2 + c m
Distance run: 1.3*3²/2 = 5.85 m
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