Answer:
Therefore, the revolutions that each tire makes is:

Explanation:
We can use the following equation:
(1)
The angular acceleration is:



and the initial angular velocity is:



Now, using equation (1) we can find the revolutions of the tire.

Therefore, the revolutions that each tire makes is:

I hope it helps you!
Answer:

Explanation:
<u>Given:</u>
- Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
- Distance of separation between the plates, d = 1.0 cm = 0.01 m.
- Minimum value of electric field that produces spark,

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

where,
= surface charge density of the plate of the capacitor =
.
= magnitude of the charge on each of the plate.
= surface area of each of the plate =
= electrical permittivity of free space, having value = 
For the minimum value of electric field that produces spark,

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.
That waves travel faster than the wave lenght!
Meiosis creates the gamete cells, because these cells are used in reproduction they only have 26 chromosomes two of these cells join together to make a full 46 chromosomes.
Hope this helps! :)
Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is
The temperature of water is
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)

