Answer:
b. Decreases
Explanation:
The total resistance of a series circuit is equal to the sum of the individual resistances:
(1)
Therefore, as we add more lamps, the total resistance increases (because we add more positive tems in the sum in eq.(1).
The current in a circuit is given by Ohm's law:
where V is the voltage provided by the power source and 
is the total resistance. We notice that the current, I, is inversely proportional to the total resistance: therefore, when more lamps are added to the series circuit, the total resistance increases, and therefore the current in the circuit decreases.
Answer:
A) c₁ = m, c₂ = m/s
B) c₁ = m/s²
C) c₁ = m/s²
D) c₁ = m/s c₂ = °
E) c₁ = m/s , c₂ = /s
Explanation:
A) x = c₁ + c₂t
⇒m = m + (m/s)s (Only same units can be added)
⇒m = m
So, c₁ = m, c₂ = m/s
B) x = 0.5c₁t²
⇒m = 0.5 (m/s²)s²
⇒m = m
So, c₁ = m/s²
C) v² = 2c₁x
⇒m²/s² = 2 (m/s²)m
⇒m²/s² = m²/s²
So, c₁ = m/s²
D) x = c₁ cos(c₂)t
⇒m = (m/s) cos(°)s
⇒m = m
So, c₁ = m/s c₂ = °
E) v² = 2c₁v-(c₂x)²
⇒m²/s² = 2(m/s)(m/s)-(1/s²)(m²)
⇒m²/s² =m²/s²
So, c₁ = m/s , c₂ = /s
Multiply I by V, you get power (energy dissipation per second). then you need to muliply by t to get the energy.
V^2 = vi^2 + 2as
vi = 0 in this case
a = 9.8 m/s^2
So,
V^2 = 2 * 9.8 * 490
v = 98.05 m/s
hope this helps
No each atom bonds in a different way according to what does it bonds with
