Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
And the correct answer is A) on the surface of the moon; because Newton's second law provides the explanation for the behavior of objects upon which the forces do not balance. The law states that unbalanced forces cause objects to accelerate with an acceleration that is directly proportional to the net force and inversely proportional to the mass.
So the correct answer is A) on the surface of the moon
Hope I helped.
The answer is actually True, I just took the test and it was correct.
Atomic Mass Unit is the answer
The formula for the energy in a capacitor , u in terms of q and c is q²/2c
<h3>What is the energy of a capacitor?</h3>
The energy of a capacitor u = 1/2qv where
- q = charge on capacitor and
- v = voltage across capacitor.
<h3>What is the capacitance of a capacitor?</h3>
Also, the capacitance of a capacitor c = q/v where
- q = charge on capacitor and
- v = voltage across capacitor.
So, v = q/c
<h3>
The formula for energy of the capacitor in terms of q and c</h3>
Substituting v into u, we have
u = 1/2qv
= 1/2q(q/c)
= q²/2c
So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c
Learn more about energy in a capacitor here:
brainly.com/question/10705986
#SPJ12
Answer:
Explanation:
Initial angular velocity ω₀ = 151 x 2π / 60
= 15.8 rad /s
final velocity = 0
Angular deceleration α = 2.23 rad / s
ω² = ω₀² - 2 α θ
0 = 15.8² - 2 x 2.23 θ
= 55.99 rad
one revolution = 2π radian
55.99 radian = 55.99 / 2 π no of terns
= 9 approx .
