Answer:
The tire fills up just like anything else that holds air when u pump a ball or tire up it fills all the way up cause it is a small confined space and after filling it with air the atoms of the air fill the tire up
Explanation:
1.) Ohm's law is understood as I = V/R. Given that resistance is constant, then voltage changes directly proportional to current.
2.) The more current that passes through a lightbulb, the brighter it glows. The higher the current, the higher the power, where power determines the brightness of a bulb.
3.) A bulb has a specific limit to how much power (Watts) it can handle. Going over the limit would cause the bulb to burn out.
4.) When a bulb burns out, no current will be able to pass through the filament.
Particles begin to move out of their ordered arrangement due to energy. They are unable to escape because of external pressure pushing down on the liquid.
Answer:
Explanation:
Given
Radius of A is twice of B i.e.

Also Potential of both sphere is same


thus




(b)Ratio of 
Electric Field is given by 
thus
----1
----2
Divide 2 by 1


Two resistors in series are often called a 'voltage divider', because the
total voltage divides in proportion to the resistances.
The total resistance in the string across the battery is (11 + 6) = 17 ohms.
-- The full battery voltage appears across 17 ohms.
-- The voltage across the 11-ohms is (11/17) of the battery, and
-- the voltage across the 6-ohms is (6/17) of the battery.
(6/17) x (B) = 9 volts
Multiply each side by (17/6) : B = (9 volts) x (17/6) =<em> 25.5 volts </em>.
By the way, in case you care or are asked . . .
-- The current in the whole series loop is B/R = 25.5 / 17 = 1.5 Amperes
-- The power drawn from the battery is B²/R = (25.5)²/17 = 38.25 watts
-- The power dissipated by the 6-ohm resistor is V²/R = 9²/6 = 13.5 watts
-- The power dissipated by the 11-ohm resistor is I²R = (1.5)² (11) = 24.75W
-- (Check: 13.5W + 24.75W = 38.25W yay! )
-- If they're just composition units hanging out in the air, then both resistors
are getting quite warm.