D) The speed of a wave slows as it travels at different speed in different media.
Answer;
A plain
Explanation;
A contour line connects points of the same elevation. Contour lines are usually curves. Closed contours represent hills.
Contour lines can not cross since they represent different elevations. A contour interval is the difference in elevation between one contour and an adjacent contour.
<u>The two ways to find acceleration in non uniform motion are as follows:</u>
<u>Explanation:</u>
Non-uniform acceleration comprises the most common description of motion. Acceleration refers to the rate of changes of velocity per unit time. Basically, it implies that acceleration changes during motion. This variety can be communicated either as far as position (x) or time (t).
Accordingly, non-uniform acceleration motion can be carried out in 2 ways:
Calculus analysis is general and accurate, but limited to the availability of speed and acceleration expressions. It is not always possible to get the expression of motion attributes in the form "x" or "t". On the other hand, the graphic method is not accurate enough, but it can be used accurately if the graphic has the correct shapes.
The use of calculations involves differentiation and integration. Integration enables evaluation of the expression of acceleration of speed and expression of movement at a distance. Similarly, differentiation allows us to evaluate expression of speed position and expression speed to acceleration.
Answer: q = 2.781e-9C = 2.781nC
E=200C
Explanation:
E = Qd/(2πEor^3)
Where
E=Electric field intensity
Q=Charge
d=distance between the dipole=0.008m
Eo=permitivitty
400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)
Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008
q = 2.781e-9C = 2.781nC
b)
Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:
E = kq*2sin θ/r^2
= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2
= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2
=200 C
Λ
![= 550 * 10^{-9}\] m. D = 3.50*10^-6 m.](https://tex.z-dn.net/?f=%20%3D%20550%20%2A%2010%5E%7B-9%7D%5C%5D%20m.%0A%0AD%20%3D%203.50%2A10%5E-6%20m.)
Then you need to find x when n=1
or
≈θ for θ<<1 , so θ=
Then you can find rhe distance of<span> bright fringe from the center:</span>
