Well there has to have a altering linear position
Answer:
The angular acceleration of the pencil<em> α = 17 rad·s⁻²</em>
Explanation:
Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:
τ = I α (1)
W r = I α (2)
The weight is that the pencil has is,
sin 10 = r / (L/2)
r = L/2(sin(10))
The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:
I = 1/3 M L²
Thus,
mg(L / 2)sin(10) = (1/3 m L²)(α)
α(f) = 3/2(g) / Lsin(10)
α = 3/2(9.8) / 0.150sin(10)
<em> α = 17 rad·s⁻²</em>
Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>
Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
