The resistance of the lamp is apparently 50V/2A = 25 ohms.
When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.
In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.
The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.
The power delivered by the 200V mains is (200V) x (2A) = 400 watts.
The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.
The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.
Together, they add up to the 400 watts delivered by the mains.
CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
<h2>
Density of the unknown liquid is 771.93 kg/m³</h2>
Explanation:
An empty graduated cylinder weighs 55.26 g
Weight of empty cylinder = 55.26 g = 0.05526 kg
Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³
Weight of cylinder plus liquid = 92.39 g = 0.09239 kg
Weight of liquid = 0.09239 - 0.05526
Weight of liquid = 0.03713 kg
We have
Mass = Volume x Density
0.03713 = 48.1 x 10⁻⁶ x Density
Density = 771.93 kg/m³
Density of the unknown liquid is 771.93 kg/m³
i think it’s B. sorry if i’m wrong
Fg=m•g || IE: Weight = mass x gravity
Therefore, the relationship are as follows:
mass and gravity are inversely proportional
mass and weight are directly proportional
weight and gravity are directly proportional
