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Mass that the block A is Ma = 10kg
Mass the the bowl is Mp = 5kg
Coefficient of restitution in between the block and also that of the bowl is e=0.75
Unstretched size of the spring is 0.6m
From the conservation of energy, us have 4 motion the block A native (1) to (2)
Then accordingly, us will have actually the action by step calculation in the attached image below.
A 15 cm × 15 cm circuit plank dissipating 20 W of power uniformly is cooled through air, i beg your pardon approached the circuit board at 20C w
A section of highway has a speed-flow partnership of the kind . What is the volume of the highway section, the speed at capac
The volume of the highway section = 400 vehicle/hr
The volume of the highway section = 40 km/hr
The density when the highway is at one-quarter the the capacity = 13.39 vehicle/km
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A fluid drug, with the viscosity and also density the water, is to be administered v a hypodermic needle. The within diameter o
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(a) The maximum volume circulation rate because that which the flow will it is in laminar is 0.0190 cubic meter every second
(b) The press drop required to supply the maximum flow rate is 148962.96 Pascal
(c) The corresponding wall shear tension is 7600 Pascal
Reynolds number = 2299, density of water = 1000kg/m^3, diameter the needle = 0.27mm = 0.00027m, length of needle = 50mm = 0.05m, viscosity the water = 0.00089kg/ms, area = 0.05m × 0.05m = 0.0025m^2, coefficient of friction = 64 ÷ Reynolds number = 64 ÷ 2299 = 0.028
Velocity = (Reynolds number × viscosity) ÷ (density × diameter) = (2299 × 0.00089) ÷ (1000 × 0.00027) = 2.046 ÷ 0.27 = 7.58m/s
(a) preferably volume circulation rate = velocity × area of needle = 7.58 × 0.0025 = 0.0190 cubic meter every second
(b) push drop = ( coefficient the friction × length × thickness × velocity^2) ÷ (2 × diameter) = (0.028 × 0.05 × 1000 × 7.58^2) ÷ (2 × 0.00027) = 80.44 ÷ 0.00054 = 148962.96 Pascal
(c) wall surface shear tension = (density × volume circulation rate) ÷ area = (1000 × 0.0190) ÷ 0.0025 = 7600 Pascal
londonchinatown.org: inherent width in the emissions line: 9.20 × 10⁻¹⁵ m or 9.20 fm
size of the photon emitted: 6.0 m
The emitted wavelength is 589 nm and also the change time is ∆t = 20 ns.
Recall the Heisenberg\"s skepticism principle:-
∆t∆E ≈ h ( Planck\"s Constant)
The shift time ∆t coincides to the power that is ∆E
The matching uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To uncover the matching spread in wavelength and also hence the line broad ∆λ, we have the right to differentiate