Answer:
C.30°
Explanation:
Given that compound swivel base is set on 60° at the lathe center line index.
We need to find reading on cross slide index
We know that relationship between center line index and cross slide index in angle 2∝=β
Where ∝ Angle of swivel and β is the reading on cross line index.
So by using above relationship between center line index and cross slide index
2∝=60°⇒∝=30°
So our option is C.
Answer: exceed 42,000 psi. Assume that the modulus of elasticity of the titanium is 16,500,000 psi
Explanation: sorry if im wrong and have a wonderful day!!!!
Answer: (a). E = 3.1656×10³⁴ √k/m
(b). f = 9.246 × 10¹² Hz
(c). Infrared region.
Explanation:
From Quantum Theory,
The energy of a proton is proportional to the frequency, from the equation;
E = hf
where E = energy in joules
h = planck's constant i.e. 6.626*10³⁴ Js
f = frequency
(a). from E = hf = 1 quanta
f = ω/2π
where ω = √k/m
consider 3 quanta of energy is lost;
E = 3hf = 3h/2π × √k/m
E = (3×6.626×10³⁴ / 2π) × √k/m
E = 3.1656×10³⁴ √k/m
(b). given from the question that K = 15 N/m
and mass M = 4 × 10⁻²⁶ kg
To get the frequency of the emitted photon,
Ephoton =hf = 3h/2π × √k/m (h cancels out)
f = 3h/2π × √k/m
f = 3h/2π × (√15 / 4 × 10⁻²⁶ )
f = 9.246 × 10¹² Hz
(c). The region of electromagnetic spectrum, the photon belongs to is the Infrared Spectrum because the frequency ranges from about 3 GHz to 400 THz in the electromagnetic spectrum.
Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
![Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]](https://tex.z-dn.net/?f=Q%20-%20%5Cdot%7BW%20%7D%20%3D%20%5Cdot%7Bm%7D%5Cleft%20%5B%20%5Cleft%20%28h_%7B2%7D-h_%7B1%7D%20%20%5Cright%20%29%2B%5Cdfrac%7Bv_%7B2%7D%5E%7B2%7D-%20v_%7B1%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20g%28z_%7B2%7D-z_%7B1%7D%29%5Cright%20%5D)
Which gives;
![560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]](https://tex.z-dn.net/?f=560%20-%20%5Cdot%7BW%20%7D%20%3D%208%5Cleft%20%5B%20%5Cleft%20%282599.2418-3755.39%20%20%5Cright%20%29%2B%5Cdfrac%7B15%5E%7B2%7D-%2060%5E%7B2%7D%7D%7B2%7D%20%5Cright%20%5D)
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
= -22749.1856 - 560 = -23309.1856 kJ
= 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
Answer:
Explanation:
For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.
