Answer:
643N.m
Explanation:
From this question we have:
Mass flow = 4kg/s
Velocity V = 400m/s
Rotation N = 1500rev/min
We get the relative velocity at exit to be:
V2 = V - r2w
400-0.5x [(2*π*1500)/60]
= 400-78.5
= 321.5m/s
Then we have to calculate the frictional torque My
Mt = Mr2 x V2
= 4x0.5x321.5
= 643Nm
From the calculations above, we get the frictional torque M on the tube to be 643Nm.
Answer:
A)
B) 
Explanation:
Using the free body diagram and according to Newton's first law, we have:
A) Solving (1) for T:
B) Solving (2) for F:
The divisions within an atom's shell are called subshells. This means that each shell consists of several subshells that are made up of orbitals. Each orbital consists of 1 or 2 electrons. The outermost shell of an atom is what we call the valence electrons, and they are what participate in chemical bonding.
Answer:
work done = ( force × displacement)
(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.
work done by gravity = (250 × 10) = 2500 joule
(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.
Answer:
The focal length of eye piece is 6.52 cm.
Explanation:
Given that,
Angular Magnification of the microscope M = -46
the distance between the lens in microscope L= 16 cm
The focal length of objective f₀ = 1.5 cm
Normal near point N = 25 cm
Have to find focal length of eye piece f ₙ =?
The angular magnification is given by
M ≈ - (L-fₙ)N/f₀fₙ
Rearranging for fₙ
fₙ =L(1 - Mf₀/N)⁺¹
=18/2.76
fₙ = 6.52 cm
The focal length of eye piece is 6.52 cm.
