Answer:
The bottom of the sea is 25 m below sea level.
Explanation:
Given data
Mass = 6.1 × 

We know that Buoyant force on the tank is equal to gravity force of the tank.



1020 ×
= 6.1 × 
= 598039.21 
We know that
= W × L × H
598039.21 = 300 × 80 × H
H = 25 m
Therefore the bottom of the sea is 25 m below sea level.
Explanation:
c. if the vector is oriented at 0° from the X -axis.
1). The equation is: (speed) = (frequency) x (wavelength)
Speed = (256 Hz) x (1.3 m) = 332.8 meters per second
2). If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.
If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.
If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.
3). The equation is: Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)
Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.
That's ' 200 k Hz ' .
Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised. They broadcast on several different frequencies,
and one of them is 198 kHz !
An apple falling to the ground is not an example of centripetal acceleration.
let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.
the mathematical formula for potential is 
for positive charges the potential is positive and is negative for negative charges.
the formula for electric field is given as-
for positive charges,the line filed is away from it and for negative charges the filed is towards it.
we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.
here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.
now let the two charges of same nature.let these are positively charged.
here we can not find a point between two charges and on the line joining two charges where the potential is zero.
but at the mid point of the line joining two charges the filed is zero.