Answer:
A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?
Since it is restricted at both ends, λ/2 = length of string
λ/2 = 1.5m
λ = 1.5*2 = 3m
The answer to the question is sound
Answer:
71 rpm
Explanation:
Given that:
Angular momentum (L) = 0.26
Diameter = 25cm = 0.25 cm
Radius, r = (d/2) = 0.125m
Mass = 5.6 kg
Moment of inertia (I) = 2mr² / 5
I = (2 * 5.6 * 0.125^2) / 5
= 0.175
= 0.175 / 5
= 0.035 kgm²
Angular speed (w) ;
w = L / I
w = 0.26 / 0.035
= 7.4285714
= 7.429 rad/s
w = (7.429 * 60/2π)
w = 445.74 / 2π rpm
w = 70.941724
Angular speed = 70.94 rpm
= 71 rpm