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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E=the sound waves from the trombones have slightly different velocity
Answer:
Option B

Explanation:
Given information
Radius of container, r=12cm=12/100=0.12m
Angular velocity= 2 rev/s, converted to rad/s we multiply by 2π
Angular velocity, 
We know that speed, 
Centripetal acceleration,
and substituting
we obtain that

Substituting \omega for 12.56637061 and r for 0.12

Rounded off, 
Answer:
Mass of the cart = 146 kg
Explanation:
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.
The cart accelerates at 1.4 m/s² horizontally.
Horizontal force = Fcosθ = 250 cos35° = 204.79N
We have F = ma
Substituting
204.79 = m x 1.4
m = 146.28 kg = 146 kg
Mass of the cart = 146 kg