🔒 Closed Sino papaunlock? (Coursehero)

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[Kene Seu]

nice malaking tulong to. Keep sharing ts <3

 

[Kene Seu]

pa unlock din ako sa susunod ts. ngayon wala pa naman e hehe

 

[Yimpskie]

Bookmark na kita papsi, Need kita sa modules ko haha

 

[Kenji246]

Boss paturo kung paano mag unlock di kame tinatantanan ng mga prof namen hanggat humihinga kame

 

[Oagari yo]

láρág links dali

 

[Heathcliff]

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paunlock po salamat

 

[Oagari yo]

eto

 

[YanAngAno]

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pa try paps

 

[Heathcliff]

maraming salamat po

 

[Oagari yo]

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pa try paps

 

[YanAngAno]

thanks pooo

 

[Cherprang Areekul]

Pa unlock po boss, Thanks

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[tantan0398]

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Paunlock ts.salamat

 

[Oagari yo]

wait lang bossing brb

 

[Oagari yo]

Pa unlock po boss, Thanks

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(A) 0.10562 m (negative correction)
(B) 0.0017 m (negative correction)
(C) 459.198 m

Step-by-step explanation
Given data:
Standard Length of tape, l = 30 m
Measured length on the slope, L = 459.20 m
Difference in elevation, h= 1.25 m
Standardization at temperature = 100 and pull = 50N
The measurement at temp = 150 and pull = 75N
The cross sectional area of tape, A = 6.50 mm2 = 6.5*10-6 m2
modulus of elasticity, E = 200 GPa = 2 *106 kN/m2
mass per unit length of tape, m = 0.075 kg/m
Let the coefficient of thermal expansion of the tape = 12*10-6/0C

(A) The Total correction per tape length:
Corrections have to be worked out for temperature, pull and sag of the tape;
Correction for temperature = lαT = 30 * (12*10-6) * (15 - 10) = 0.0018 m
Correction for pull = AEΔPl = [(75 - 50) * 30] / (6.5*10-6*200*109) = 0.00058 m
Weight of the tape = 0.075(kg/m)*9.81*30 = 22.0725N
Correction for sag = l(wl)2/24P2 = 30*(22.0725)2/(24*752) = 0.108 m
So, Total correction per tape length = 0.0018+0.00058-0.108 = -0.10562 m (correction is negative)

(B) The correction for slope:
Correction for slope = 2∗Lh2 = 1.252/(2*459.20) = 0.0017 m, this correction is always negative.

(C) The horizontal distance:
Using pythagoras theorem, Horizontal distance,D = L2−h2 = (459.20)2−(1.25)2 = 459.198 m

 

[Oagari yo]

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Paunlock ts.salamat

 

[thelastswordsma]

paano yan lods ?

 

[thelastswordsma]

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sana lods

 

[Cherprang Areekul]

(A) 0.10562 m (negative correction)
(B) 0.0017 m (negative correction)
(C) 459.198 m

Step-by-step explanation
Given data:
Standard Length of tape, l = 30 m
Measured length on the slope, L = 459.20 m
Difference in elevation, h= 1.25 m
Standardization at temperature = 100 and pull = 50N
The measurement at temp = 150 and pull = 75N
The cross sectional area of tape, A = 6.50 mm2 = 6.5*10-6 m2
modulus of elasticity, E = 200 GPa = 2 *106 kN/m2
mass per unit length of tape, m = 0.075 kg/m
Let the coefficient of thermal expansion of the tape = 12*10-6/0C

(A) The Total correction per tape length:
Corrections have to be worked out for temperature, pull and sag of the tape;
Correction for temperature = lαT = 30 * (12*10-6) * (15 - 10) = 0.0018 m
Correction for pull = AEΔPl = [(75 - 50) * 30] / (6.5*10-6*200*109) = 0.00058 m
Weight of the tape = 0.075(kg/m)*9.81*30 = 22.0725N
Correction for sag = l(wl)2/24P2 = 30*(22.0725)2/(24*752) = 0.108 m
So, Total correction per tape length = 0.0018+0.00058-0.108 = -0.10562 m (correction is negative)

(B) The correction for slope:
Correction for slope = 2∗Lh2 = 1.252/(2*459.20) = 0.0017 m, this correction is always negative.

(C) The horizontal distance:
Using pythagoras theorem, Horizontal distance,D = L2−h2 = (459.20)2−(1.25)2 = 459.198 m

Thanks