Hi!
SI units are physical measurements which will be in the form of kilograms, second, kelvin, metres, etc.
Since kilograms measure the weight of an object, it is out. Miles and feet are not SI units, so they are also out. This only leaves one answer left!
Hopefully, this helps! =)
True, scientists often talk to each other to figure out if their results were similar and what they could have done better.
Although, talking to other scientists does have risks, other scientists could copy your work and further better it.
So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol
Correct option B
Transverse waves are those waves whose particles vibrate perpendicular to the direction of wave.
Hope This Helps You!
Answer:
Speed is 0.08 m/s.
Explanation:
Given the distance that the bird flies = 3.7 meters
The time is taken by the bird to fly the 3.7 meters = 46 seconds
We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.
Below is the formula to find the speed.
Speed = Distance / Time
Now insert the given value in the formula.
Speed = 3.7 / 46 = 0.08 m/s
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)

(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)



The charge chould be moved to infinity