I really don’t know good luck
Answer:
Explanation:
Work, U, is equal to the force times the distance:
U = F · r
Force needed to lift the weight, is equal to the weight: F = W = m · g
so:
U = m · g · r
= 20.4kg · 9.81 
· 1.50m
= 35.316 
= 35.316 W
Answer:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Explanation:
Calculation to estimate the upper and lower bounds of the modulus of this composite.
First step is to calculate the maximum modulus for the combined material using this formula
Modulus of Elasticity for mixture
E= EcuVcu+EwVw
Let pug in the formula
E =( 110 x 0.40)+ (407 x 0.60)
E=44+244.2 GPa
E=288.2GPa
Second step is to calculate the combined specific gravity using this formula
p= pcuVcu+pwTw
Let plug in the formula
p = (19.3 x 0.40) + (8.9 x 0.60)
p=7.72+5.34
p=13.06
Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness
UPPER BOUNDS
Using this formula
Upper bounds=E/p
Let plug in the formula
Upper bounds=288.2/13.06
Upper bounds=22.07 GPa
LOWER BOUNDS
Using this formula
Lower bounds=EcuVcu/pcu+EwVw/pw
Let plug in the formula
Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3
Lower bounds=(44/8.9)+(244.2/19.3)
Lower bounds=4.94+12.65
Lower bounds=17.59 GPa
Therefore the Estimated upper and lower bounds of the modulus of this composite will be:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Answer:
square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.
Required:
a. What is the nominal maximum tensile stress on the bar?
b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?
Answer: 0.95 inches
Explanation:
A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.
The length= 64 inches
Ends are fixed Le= 64/2 = 32 inches
Factor Of Safety (FOS) = 3. 0
E= 10.6× 10^6 ps
σy= 4000ps
The square cross-section= ia^4/12
PE= π^2EI/Le^2
6500= 3.142^2 × 10^6 × a^4/12×32^2
a^4= 0.81 => a=0.81 inches => a=0.95 inches
Given σy= 4000ps
σallowable= σy/3= 40000/3= 13333. 33psi
Load acting= 6500
Area= a^2= 0.95 ×0.95= 0.9025
σactual=6500/0.9025
σ actual < σallowable
The dimension a= 0.95 inches
