Answer:
3.03 m/s²
Explanation:
m₂ > m₁, so m₁ will move up, m₂ will move down, and m₃ will move to the right.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₂ − m₂g = m₂(-a)
T₂ = m₂g − m₂a
Sum of the forces on m₃ in the y direction:
∑F = ma
N − m₃g = 0
N = m₃g
Sum of the forces on m₃ in the x direction:
∑F = ma
T₂ − T₁ − F = m₃a
T₂ − T₁ − Nμ = m₃a
Substituting and solving for acceleration:
(m₂g − m₂a) − (m₁g + m₁a) − (m₃g)μ = m₃a
m₂g − m₂a − m₁g − m₁a − m₃gμ = m₃a
g (m₂ − m₁ − m₃μ) = (m₁ + m₂ + m₃) a
a = g (m₂ − m₁ − m₃μ) / (m₁ + m₂ + m₃)
Given m₁ = 4 kg, m₂ = 12 kg, m₃ = 6 kg, and μ = 0.2:
a = 9.81 (12 − 4 − 6×0.2) / (4 + 12 + 6)
a = 3.03 m/s²
Centripetal acceleration is given by:
a = v²/r
a = centripetal acceleration, v = speed of cylinder at surface, r = radius of cylinder
This equation relates the speed of the cylinder at its surface to its angular velocity:
v = rω
v = speed at surface, r = radius, ω = angular velocity
Make a substitution:
a = (rω)²/r
a = rω²
Given values:
a = 9.81m/s², r = (5.0mi)/2 = 2.5mi = 4023m
Plug in and solve for ω:
9.81 = 4023ω²
ω = 0.049rad/s
Answer:
There are several possibilities why a battery can feel hot or heat up. In case of an external short circuit: this can happen when the battery is short circuited i.e. in a purse, or in a drawer. ... Then there is also the possibility of a short circuit, forced charging and heating up the battery.
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