B) a rock being tossed high into the air
The electrical force acting on a charge q immersed in an electric field is equal to

where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
<h2>~<u>Solution</u> :-</h2>
- Here, the <u>moment arm</u> is defined as follows;
The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.

(a) The spring stiffness constant of the spring is 18,392 N/m.
(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.
<h3>Kinetic energy of the car</h3>
The kinetic energy of the car is calculated as follows;
K.E = ¹/₂mv²
K.E = ¹/₂ x 950 x 22²
K.E = 229,900 J
<h3>Stiffness constant of the spring</h3>
The stiffness constant of the spring is calculated as follows;
K.E = U = ¹/₂kx²
k = 2U/x²
k = (2 x 229,900)/(5)²
k = 18,392 N/m
<h3>Force exerted on the spring</h3>
F = kx
F = 18,392 x 5
F = 91,960 N
<h3>Time of impact</h3>
F = mv/t
t = mv/F
t = (950 x 22)/(91960)
t = 0.23 s
Learn more about spring constant here: brainly.com/question/1968517
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The noble gases have eight valence electrons and as a result are stable.
If an atom consists of 8 valence electrons, they have a full octet, and do not need to bond, which makes them "happy".