Gravity largely depends on the comparison of two objects; it's why you have the equation F= (GMm)/r^2. On Earth, you have different altitudes that, with the formula, will give different results for gravity because the radius is different everywhere. This difference on calculations, however, are seen to be miniscule. We know gravity as 9.81 m/s^2 but it might be different by thousandths or hundreds of thousandths of a decimal.
If you are talking about ocean waves crashing into each other, they would probably mostly cancel out with just a bit of motion left over. If you are talking about things like frequency and amplitude, overlapping waves would combine and amplify or suppress each other, depending on their direction, position, frequency and amplitude. If the two waves complement each other, they amplify; if they conflict with each other, they are suppressed.
Answer: Here is the complete question:
A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 103 N/C. If the ball is in equilibrium when the string makes a 30 angle with the vertical, what is the net charge on the ball?
Answer: The charge on the ball is 5.71 × 10^-4 C
Explanation:
Please see the attachments below
Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft × 4 sorings
K = 100 lb/ft
Amplitude of the microscope ![\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]](https://tex.z-dn.net/?f=%5Cfrac%7BX%7D%7BY%7D%3D%20%5B%5Cfrac%7B1%2B2%20%5Cepsilon%20%28%5Comega%2F%20W_n%29%5E2%7D%7B%281-%28%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%29%5E2%2B%282%20%5Cepsilon%20%20%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%7D%5D)
where;


= 
= 4.0124
replacing them into the above equation and making X the subject of the formula:



Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Answer:
The true course:
north of east
The ground speed of the plane: 96.68 m/s
Explanation:
Given:
= velocity of wind = 
= velocity of plane in still air = 
Assume:
= resultant velocity of the plane
= direction of the plane with the east
Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

Let us find the direction of this resultant velocity with respect to east direction:

This means the the true course of the plane is in the direction of
north of east.
The ground speed will be the magnitude of the resultant velocity of the plane.

Hence, the ground speed of the plane is 96.68 km/h.