No, it only does when entering an atmosphere
Answer: Scanning Tunneling Microscope
Explanation:
The answer is, of course, the greatest Star Trek fan art imaginable: images literally built out of individual atoms. The images are the work of IBM scientists who created the unique artwork with a two-ton machine called the Scanning Tunneling Microscope that moves single atoms across a tiny piece of copper.
Answer:
(a) 
(b) 
Explanation:
<u>Given:</u>
= The first temperature of air inside the tire = 
= The second temperature of air inside the tire = 
= The third temperature of air inside the tire = 
= The first volume of air inside the tire
= The second volume of air inside the tire = 
= The third volume of air inside the tire = 
= The first pressure of air inside the tire = 
<u>Assume:</u>
= The second pressure of air inside the tire
= The third pressure of air inside the tire- n = number of moles of air
Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.
Using ideal gas equation, we have
Part (a):
Using the above equation for this part of compression in the air, we have
Hence, the pressure in the tire after the compression is .
Part (b):
Again using the equation for this part for the air, we have
Hence, the pressure in the tire after the car i driven at high speed is 
.
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
