Components connected in series are connected along a single path, so the same current flows through all of the components. If the light bulbs are connected in parallel, the currents through the light bulbs combine to form the current in the battery, while the voltage drop is across each bulb and they all glow.
The answer to your question is true.
Answer:
ac = 3.92 m/s²
Explanation:
In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,
Frictional Force = Centripetal Force
where,
Frictional Force = μ(Normal Force) = μ(weight) = μmg
Centripetal Force = (m)(ac)
Therefore,
μmg = (m)(ac)
ac = μg
where,
ac = magnitude of centripetal acceleration of car = ?
μ = coefficient of friction of tires (kinetic) = 0.4
g = 9.8 m/s²
Therefore,
ac = (0.4)(9.8 m/s²)
<u>ac = 3.92 m/s²</u>
Distance, since distance represents how far something has travelled, which would be in our case 2.5m.
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
