The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.
The current flowing through the lamp is given by Ohm's law:
where
V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.
The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.
Electron;Neutron is the correct answer.
Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3
Answer:
Nuclear Forces
Explanation:
The type of force that holds the nucleus of an atom together is called Nuclear Forces.
