Answer:
None
Explanation:
An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:
- Verbal scale use of words to represent scale information on the map. The distance or linear units are used for depicting this scale on the map. For example: 1 inch = 1 Kilo meter.
- Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio. Example: 1: 10,000 or 1/10,000
- In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given. Example: 1:500
- Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map. Example: 1:25,000
- A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.
I hope you find this information useful and interesting! Good luck!
Because we can reproduce with just one and if more eggs were released, the female reproductive life (which is 40 years) would be shorter and so there would be less time for every female human to reproduce since the amount of eggs is limited. (Also fraternal twins come from 2 eggs released at the same time)
<span>a driver's foot off the accelerator and on the break pedal. C</span>
Answer:
Electric field, E = 40608.75 N/C
Explanation:
It is given that,
Mass of electrons, 
Initial speed of electron, u = 0
Final speed of electrons, 
Distance traveled, s = 6.3 cm = 0.063 m
Firstly, we will find the acceleration of the electron using third equation of motion as :
Now we will find the electric field required in the tube as :
E = 40608.75 N/C
So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.
