Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
Answer:
A. It is always a positive force
Explanation:
Hooke's law describes the relation between an applied force and extension ability of an elastic material. The law states that provided the elastic limit, e, of a material is not exceeded, the force, F, applied is proportional to the extension, x, provided temperature is constant.
i.e F = - kx
where k is the constant of proportionality, and the minus sign implies that the force is a restoring force.
The applied force can either be compressing or stretching force.
Answer:
Explanation:
Using the pythagoras theorem, the displacement is expressed as;
d² = x²+y²
y = 36m (north)
x = 20m east
Substitute;
d² = 36²+20²
d² = 1296+400
d² = 1696
d = √1696
d = 41.18m
For the direction;
theta = tan^-1(y/x)
theta = tan^-1(36/20)
theta = tan^-1(1.8)
theta = 60.95°
Hence the magnitude is 41.18m and the direction is 60.95°
