Answer:
I got iron
Explanation:
on my plato test
If the roller at A and the pin at B can support a load upto 4kN and 8kN, repsectively, determine the maximum intensity of the distributed load w, measured in kN/m, so that failure of the supports does not occur.
Solution :
Taken moment about point B;
[tex]$\sum M_B = 0$[/tex]
[tex]$\left[(w)(4)(2)-N_A \sin 30^\circ (3)(\sin 30^\circ)-N_A \cos 30^\circ(3 \cos 30^\circ +4) \right] =0$[/tex]
[tex]$N_A = 1.2376w$[/tex]
[tex]$\sum F_x = 0$[/tex]
[tex]$N_A \sin 30^\circ - B_x = 0$[/tex]
[tex]$B_x = 1.2376w \times \sin 30^\circ$[/tex]
[tex]$B_x = 0.6188w$[/tex]
[tex]$\sum F_y = 0$[/tex]
[tex]$B_y+N_A \cos^\circ - (w \times 4) = 0$[/tex]
[tex]$B_y+(1.2376w) \cos 30^\circ - (w \times 4)=0 $[/tex]
[tex]$B_y = 2.9282w$[/tex]
Now calculating the resultant force at B,
[tex]$F_B= \sqrt{B_x^2+B_y^2}$[/tex]
[tex]$F_B= \sqrt{(0.6188w)^2+(2.9282w)^2}$[/tex]
= 2.9929w
For no failure,
[tex]$N_A<4 \ kN $[/tex]
2.9929 w < 4 kN
[tex]$w < 3.232 \ kN/m$[/tex]
And,
[tex]$F_B < 8 \ kN$[/tex]
2.9929 w < 8 kN
w < 2.673 kN/m
For the safe operation, w = 2.673 kN/m
Compare and contrast the roles of agricultural and environmental scientists.
Answer:
HUHHHHHH BE SPECIFIC CHILE
Explanation:
ERM IRDK SORRY BOUT THAT
If 1 inch equals 10 feet, what would the measured distance be if the line scaled 1 1/2 inches?
A. 10 feet
b. 15 feet
c. 2 feet
d. 25 feet
Answer: B. 15 feet
Explanation:
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R14B, (6) G1B, (7) M8C3, (8) P1NSE, and (9) RL1.
(a) Determine the normal times in TMUs for these motion elements.
(b) What is the total time for this work element in sec
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
1. The construction process begins with which of the following stages?
a) establishing the foundation
b) enclosing the building
c) site preparation
d) installing underground utilities
Answer:
c) site preparation
Explanation:
A construction process can be defined as a series of important physical events (processes) that must be accomplished during the execution of a construction project.
Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.
This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.
Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.
Additionally, construction costs can be defined as the overall costs associated with the development of a built asset, project or property. The construction costs is classified into two (2) main categories and these are; capital and operational costs.
Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distance that a 1 (GHz) wave travels in graphite such that its field intensity is reduced by 30 (dB).
Answer:
the answer is below
Explanation:
a) The conductivity of graphite (σ) is calculated using the formula:
[tex]\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}[/tex]
where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012
Substituting:
[tex]\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m[/tex]
b) f = 1 GHz = 10⁹ Hz.
[tex]\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm[/tex]
1) I love to swim. 2) A few years ago, my new year's resolution was to become a faster swimmer. 3) First, I started eating better to improve my overall health. 4) Then, I created a training program and started swimming five days a week. 5) I went to the pool at my local gym. 6) To measure my improvement, I tried to count my laps as I was swimming, but I always got distracted and lost track! 7) It made it very hard for me to know if I was getting faster. 8) This is a common experience for swimmers everywhere. 9) We need a wearable device to count laps, calories burned, and other real-time data. Summarey of the story
Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kept at a temperature of 80.6 °F year round using a natural gas heater. If you were to completely drain the pool and refill the pool with 50°F water, (blank) GJ (giga-Joules) of energy are required to to heat the water back to 80.6 °F. Note: The specific heat capacity of water is 4182 J/kg ⋅°C. The cost of natural gas per GJ is $2.844. It costs $ (blank) to heat the pool (to the nearest dollar).
Answer:
[tex]95.914\ \text{GJ}[/tex]
[tex]\$272.78[/tex]
Explanation:
m = Mass of water = 749511.5 kg
c = Specific heat of water = 4182 J/kg ⋅°C
[tex]\Delta T[/tex] = Change in temperature = [tex]80.6-50=30.6^{\circ}\text{F}[/tex]
Cost of 1 GJ of energy = $2.844
Heat required is given by
[tex]Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}[/tex]
Amount of heat required to heat the water is [tex]95.914\ \text{GJ}[/tex].
Cost of heating the water is
[tex]95.914\times 2.844=\$272.78[/tex]
Cost of heating the water to the required temperature is [tex]\$272.78[/tex].