A complex entity involving the Earth's biosphere, atmosphere, oceans, and soil; the totality constituting a feedback or cybernetic system which seeks an optimal physical and chemical environment for life on this planet
Answer:
a. N = 2.49W b. 0.40
Explanation:
a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?
Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s
Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m
The rider's weight W = mg = 9.8m
The ratio of the normal force to the rider's weight is
N/W = 24.43m/9.8m = 2.49
So the normal force expressed in term's of the rider's weight is
N = 2.49W
b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?
The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.
Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.
So, the normal force equals
N = F/μ = W/μ = mg/μ = mrω²
μ = mg/mrω²
= W/N
= 9.8m/24.43m
= 0.40
5,10,15,20,25,30, that's how much it should have been
Answer:
option (E) 1,000,000 J
Explanation:
Given:
Mass of the suspension cable, m = 1,000 kg
Distance, h = 100 m
Now,
from the work energy theorem
Work done by the gravity = Work done by brake
or
mgh = Work done by brake
where, g is the acceleration due to the gravity = 10 m/s²
or
Work done by brake = 1000 × 10 × 100
or
Work done by brake = 1,000,000 J
this work done is the release of heat in the brakes
Hence, the correct answer is option (E) 1,000,000 J
Correct option: A
An object remains at rest until a force acts on it.
As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.
