I think that this is false but I am not sure
Answer:
Explanation:
a )
We shall apply the concept of impulse .
Impulse = force x time = change in momentum
= 5 x 4 = 2 ( V - 3 ) , where V is final velocity of the object
20 = 2V - 6
V = 13 m /s
b )
Impulse applied = - 7 x 4 = - 28 kg m/s ( negative as direction of force is opposite motion )
If v be the final velocity
2 x 3 - 28 = 2 v ( initial momentum - change in momentum = final momentum )
- 22 = 2v
v = - 11 m /s
object will move with 11 m /s in opposite direction .
Answer:
The electric potential is approximately 5.8 V
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero
Explanation:
The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:
(1)
where is the charge of the particle, the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:
Substituting the values , and we obtain:
The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.
The answer is A. <span>It showed that radiometric dating could be used to accurately determine the age of fossils.</span>
Answer:
12.14 cm
Explanation:
mass, m = 15.5 kg
frequency, f = 9.73 Hz
maximum amplitude, A = 14.6 cm
t = 1.25 s
The equation of the simple harmonic motion
y = A Sin ωt
y = A Sin (2 x π x f x t)
put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz
y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)
y = 14.6 Sin 76.38
y = 12.14 cm
Thus, the displacement of the particle from the equilibrium position is 12.14 cm.