Two everyday examples of a starting motion
1 answer:
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Answer:
α = 13.7 rad / s²
Explanation:
Let's use Newton's second law for rotational motion
∑ τ = I α
we will assume that the counterclockwise turns are positive
F₁ 0 + F₂ R₂ - F₃ R₃ = I α
give us the cylinder moment of inertia
I = ½ M R₂²
α = (F₂ R₂ - F₃ R₃)
let's calculate
α = (24 0.22 - 13 0.10) 2/12 0.22²
α = 13.7 rad / s²
Answer:
B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2
C) V = 0.84m/s T = 29.92s
D) ω2 = 0.315 rad/s
Explanation:
The moment of inertia when they are standing on the edge:
where M is the mass of the merry-go-round.
I1 = 1680 kg.m^2
The moment of inertia when they are standing half way to the center:
I2 = 1120 kg.m^2
The tangencial velocity is given by:
V = ω1*R = 0.84m/s
Period of rotation:
T = 2π / ω1 = 29.92s
Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:
I1*ω1 = I2*ω2 Solving for ω2:
ω2 = I1*ω1 / I2 = 0.315 rad/s
The electric field strength is
Explanation:
The strength of the electric field produced by a single point charge is given by:
where
is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge at which the field strength is calculated
For the charge in the problem, we have:
is the charge
Therefore, the electric field strength is
Learn more about electric fields:
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