Answer:
w = 10.437 kips
deflection at 1/4 span 20.83\E ft
at mid span = 1.23\E ft
shear stress 7.3629 psi
Explanation:
area of cross section = 18*76
length of span = 32 ft
moment = 334 kips-ft
we know that
moment = load *eccentricity
334 = w * 32
w = 10.437 kips
deflection at 1/4 span
= 20.83\E ft
at mid span
shear stress
Answer:
The costs to run the dryer for one year are $ 9.03.
Explanation:
Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:
1 watt = 0.001 kilowatt
2250/45 = 50 watts per minute
45 x 365 = 16,425 / 60 = 273.75 hours of consumption
50 x 60 = 300 watt = 0.3 kw / h
0.3 x 273.75 = 82.125
82.125 x 0.11 = 9.03
Therefore, the costs to run the dryer for one year are $ 9.03.
Answer:
A transforming vechicle that could transform from a land-based vehicle to a water-based vehicle and to an air based vehicle.
Explanation:
Answer:
Q' = 8 KW.h
Q'=28800 KJ
Explanation:
Given that
Heat Q= 4 KW
time ,t = 2 hours
The amount of energy used in KWh given as
Q ' = Q x t
Q' = 4 x 2 KW.h
Q' = 8 KW.h
We know that
1 h = 60 min = 60 x 60 s = 3600 s
We know that W = 1 J/s
The amount of energy used in KJ given as
Q' = 8 x 3600 = 28800 KJ
Therefore
Q' = 8 KW.h
Q'=28800 KJ
Answer:
λ^3 = 4.37
Explanation:
first let us to calculate the average density of the alloy
for simplicity of calculation assume a 100g alloy
80g --> Ag
20g --> Pd
ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)
= 100*10^-3/(20/11.9*10^6+80/10.44*10^6)
= 10744.62 kg/m^3
now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate
total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27
= 1.12*10^23 unit cells
mass of Pd in 1 unit cell = 20/1.12*10^23
Now,
ρ_avg= mass of unit cell/volume of unit cell
ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3
λ^3 = 4.37
