Answer:
Considering the guidelines of this exercise.
The pieces produced per month are 504 000
The productivity ratio is 75%
Explanation:
To understand this answer we need to analyze the problem. First of all, we can only produce 2 batches of production by the press because we require 3 hours to set it up. So if we rest those 6 hours from the 8 of the shift we get 6, leaving 2 for an incomplete bath. So multiplying 2 batches per day of production by press we obtain 40 batches per day. So, considering we work in this factory for 21 days per month well that makes 40 x 21 making 840 then we multiply the batches for the pieces 840 x 600 obtaining 504000 pieces produced per month. To obtain the productivity ratio we need to divide the standard labor hours meaning 6 by the amount of time worked meaning 8. Obtaining 75% efficiency.
When the in-plane shear stress is at its highest, it is true that the normal stresses are equal.
Their combined value is σn+σt.
<h3>What is in-plane shear stress?</h3>
- 6.3 ksi is the maximum in-plane shear stress. 10.2 ksi is the maximum out-of-plane shear stress.
- By examining Mohr's circle, we may understand why out of plane shear stress is the highest.
- Maximum in-plane shear stress is indicated by the inner blue circle radius (6.3 ksi).
- Maximum shear stress planes are 45 degrees from the major planes.
- The principle stress is the highest stress that may be created in a plane, and the principal stress refers to the principal plane when shear stress is zero.
The complete question is:
What is true about the normal stresses when the in-plane shear stress is maximum? Select all that apply.
a) They are equal.
b) They are zero.
c) Their sum is equal to σn+σt.
d) They are maximum.
e) They are equal to shear stress.
To learn more about in-plane shear stress, refer to:
brainly.com/question/15517350
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Answer:
a mass of water required is mw= 1273.26 gr = 1.27376 Kg
Explanation:
Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:
Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L
where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation
therefore
mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )
replacing values
mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg
Answer: 133.88 MPa approximately 134 MPa
Explanation:
Given
Plane strains fracture toughness, k = 26 MPa
Stress at which fracture occurs, σ = 112 MPa
Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m
Critical internal crack length, l' = 6 mm = 6*10^-3 m
We know that
σ = K/(Y.√πa), where
112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]
112 MPa = 26 MPa / Y.√(3.142 * 0.043)
112 = 26 / Y.√1.35*10^-2
112 = 26 / Y * 0.116
Y = 26 / 112 * 0.116
Y = 26 / 13
Y = 2
σ = K/(Y.√πa), using l'instead of l and, using Y as 2
σ = 26 / 2 * [√3.142 * (6*10^-3/2)]
σ = 26 / 2 * √(3.142 *3*10^-3)
σ = 26 / 2 * √0.009426
σ = 26 / 2 * 0.0971
σ = 26 / 0.1942
σ = 133.88 MPa