Answer is OB
Answer is OB
Answer is OB
Answer:
When they are connected in series
The 50 W bulb glow more than the 100 W bulb
Explanation:
From the question we are told that
The power rating of the first bulb is 
The power rating of the second bulb is 
Generally the power rating of the first bulb is mathematically represented as

Where
is the normal household voltage which is constant for both bulbs
So

substituting values

Thus the resistance of the second bulb would be evaluated as

From the above calculation we see that

This power rating of the first bulb can also be represented mathematically as

This power rating of the first bulb can also be represented mathematically as

Now given that they are connected in series which implies that the same current flow through them so

This means that

So when they are connected in series

This means that the 50 W bulb glows more than the 100 \ W bulb
The answer is 165.3 cm³.
P1 * V1 / T1 = P2 * V2 / T2
The initial sample:
P1 = 84.6 kPa
V1 = 215 cm³
T1 = 23.5°C = 23.5 + 273 K = 296.5 K
At STP:
P2 = 101.3 kPa
V2 = ?
T2 = 273 K
Therefore:
84.6 * 215 / 296.5 = 101.3 * V2 / 273
61.34 = 101.3 * V2 / 273
V2 = 61.34 * 273 / 101.3
V2 = 165.3 cm³
Acid it is i believe........
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to

where

is the charge density

is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate: