All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.
Therefore the reaction at the centre support is equal to the weight of the board, while the support at the end has 0 reaction force.
Answer:
The advantage of the SI unit over CGS unit are:
- S.I has broader base. It has seven base units and two supplementary units.
- S.I is metric.
- S.I is coherent.
- S.I is rational, i.e. it gives one unit for one physical quantity .e g. for energy of any type i.e, mechanical or heat or electrical.There is only one unit Joule(J) but in M.K.S system unit for mechanical energy is joule.
<em><u>HOPE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>WILL</u></em><em><u> </u></em><em><u>HELP</u></em><em><u> </u></em><em><u>YOU</u></em>
Answer:
Its traveling in the +x direction
Explanation:
The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.
Answer:
Resistance = 3.35*
Ω
Explanation:
Since resistance R = ρ
whereas 
resistivity is given for two ends. At the left end resistivity is 
whereas x at the left end will be 0 as distance is zero. Thus
At the right end x will be equal to the length of the rod, so 
Thus resistance will be R = ρ
where A = π 
so,
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
