To solve the problem it is necessary to identify the equation in the manner given above.
This equation corresponds to the displacement of a body under the principle of simple harmonic movement.
Where,
PART A) Our equation corresponds to
Therefore the value of omega is equivalent to that of
From the definition we know that the period as a function of angular velocity is equivalent to
This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the 
Is equivalent to . So the maximum speed that the body can reach is,
Therefore the maximum felocity will be 5ft / s
PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then
answer:
resultant = 127.65 in the positive direction
explanation:
F1 = 50N , F2 = 40N, f3 = 55N , f4 = 60N
Fy = 50 sin 50 = 50 × -0.26 = -13
Fx = 40 cos 0 = 40×1 = 40
fx = 55 cos 25 = 55×0.99 = 54.45
Fy = 60 sin 70 = 60 × 0.77 = 46.2
resultant = -13+40+54.45+46.2 = 127.65 in the positive direction
If the separation distance is doubled, then the electric field decreases by a factor of 4.
<h3>What is the electric field strength?</h3>
We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.
We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.
Thus we know that;
E = Kq/r^2
Where;
E = electric field strength
q = magnitude of charge
r = distance of separation
Now;
E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2
E = 4 N/C
Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.
Learn more about electric field strength:brainly.com/question/15170044?
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Answer:
1.196 μm
Explanation:
D = Screen distance = 3 m
= Wavelength = 598 m
y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm
d = Slit distance
For first dark fringe
Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm
E, there is no state of matter that has no particle motion, however a solid's particles are only vibrating.
