I believe that the statement is true. <span>Investigations allow for the control of variables. Changing variables will lead you to observations that may prove your hypothesis. Hope this answers the question. Have a nice day. Please feel free to ask more questions.</span>
Out of the choices given, igniting the gas-air mixture supplies the heat for the hot reservoir in a car's engine. The correct answer is C.
C. It is answered by observation and evidence.
Good scientific explanations are defined, measurable and controllable. They can be answered by an experiment.
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>
-- Put the rod into the freezer for a while. As it cools,
it contracts (gets smaller) slightly.
-- Put the cylinder into hot hot water for a while. As it heats,
it expands (gets bigger) slightly.
-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.
-- I bet it'll fit now.
-- But be careful . . . get the rod exactly where you want it as fast
as you can. Once both pieces come back to the same temperature,
and the rod expands a little and the cylinder contracts a little, the fit
will be so tight that you'll probably never get them apart again, or even
move the rod.
