Support with three reasons the decision to use a plastic material for the package in the following
scenario.
Situation: A client has hired Jose, a materials engineer, to develop a package for an item he has begun
to market. The object needs to be mailed to customers within three days of being ordered.
Answer:
its durable. it's cheap. its recyclable
Explanation:
Plastic is made of lots of recycled materials that make it very useful and cheap.
700.0 liters of a gas are prepared at 760.0 mmHg and 100.0 °C. The gas is placed into a tank under high pressure. When the tank cools to 32.0 °C, the pressure of the gas is 20.0 atm. What is the volume of the gas?
Answer:
The volume of the gas is 11.2 L.
Explanation:
Initially, we have:
V₁ = 700.0 L
P₁ = 760.0 mmHg = 1 atm
T₁ = 100.0 °C
When the gas is in the thank we have:
V₂ =?
P₂ = 20.0 atm
T₂ = 32.0 °C
Now, we can find the volume of the gas in the thank by using the Ideal Gas Law:
[tex] PV = nRT [/tex]
[tex]V_{2} = \frac{nRT_{2}}{P_{2}}[/tex] (1)
Where R is the gas constant
With the initials conditions we can find the number of moles:
[tex] n = \frac{P_{1}V_{1}}{RT_{1}} [/tex] (2)
By entering equation (2) into (1) we have:
[tex] V_{2} = \frac{P_{1}V_{1}}{RT_{1}}*\frac{RT_{2}}{P_{2}} = \frac{1 atm*700.0 L*32.0 ^{\circ}}{100.0 ^{\circ}*20.0 atm} = 11.2 L [/tex]
Therefore, When the gas is placed into a tank the volume of the gas is 11.2 L.
I hope it helps you!
1. Discuss how products incorporate aesthetic design and why this appeals to target markets 2. Discuss how the universal design process has impacted engineering design and the impact these expectations will have on the future of product design.
Explanation:
Remember, to say a product is incorporated with aesthetic design implies that its overall appearance is designed to look beautiful to the eyes of the user/buyer. For example, a clothing company whose target market is mainly focused on women's clothing would need to take into consideration that certain colors like pink, blue, etc are attractive to women more than men. So they'll have to ensure the colors of their clothing are suitable to the needs of their target market.
The Universal Design process involves building products that can be used by a wide range of users at ease. For example, you may ask yourself: Is my product/service easily accesible to those with disabilities?
Other processes include;
Defining who the users (or universe) are of the products. Involve consumers in the design.Follow the existing standards of product designEvaluate and review your universal design methodsa cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8mm. the beam carries a uniform load of 6.5kN/m along the entire length and, in the same direction, a concentrated force of 4kN at the free end. (a) determine the max bending stress (b) determine the max transervse shear stress (c) determine the max shear stress in the beam
Answer:
a) 159.07 MPa
b) 10.45 MPa
c) 79.535 MPa
Explanation:
Given data :
length of cantilever beam = 1.5m
outer width and height = 100 mm
wall thickness = 8mm
uniform load carried by beam along entire length= 6.5 kN/m
concentrated force at free end = 4kN
first we determine these values :
Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m
Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N
A) determine max bending stress
б = [tex]\frac{MC}{I}[/tex] = [tex]\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}[/tex] = 159.07 MPa
B) Determine max transverse shear stress
attached below
ζ = 10.45 MPa
C) Determine max shear stress in the beam
This occurs at the top of the beam or at the centroidal axis
hence max stress in the beam = 159.07 / 2 = 79.535 MPa
attached below is the remaining solution
Compute the weight fraction of graphite, in a 3 wt% C Ferritic Gray cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
9.60%
Explanation:
Computation for the weight fraction of graphite
First step
Computation for the mass fraction for Wa using this formula
Wa=Cg-Co/Cg-Ca
Let plug
Wa=10-3/100-0
Wa=0.97
Computation for the mass fraction for Wg using this formula
Wg=Co-Ca/Cg-Ca
Let plug in the formula
Wg=3-0/100-0
Wg=0.03
Second step is to convert the mass fraction to Volume Fraction using this formula
Volume Fraction =[Wg/Pg÷(Wa/Pa)+(Wg/Pg)]*100
Let plug in the formula
Volume =[0.03/2.3 ÷(0.97/7.9)+(0.03/2.3)]*100
Volume=[0.0130435÷0.1227848+0.0130435]*100
Volume=[0.0130435÷0.135828]*100
Volume=0.096*100
Volume=9.60%
Therefore the weight fraction of graphite will be 9.60%
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surface emissivity of 0.95. Heat is lost from the head to the surrounding air at 25°C by convection with a convection coefficient of ???????????????? ???????? ????????????????∙???????? , and by radiation to the surrounding black walls at 15°C. Determine the total rate of heat loss. StefanBoltzmann Constant, ???????? = ????????. ???????????????? × ????????????????−???????? ???????? ????????????????∙???????????????? . (10 points)
Answer:
Hello some parts of your question is missing below is the missing part
Convection coefficient = 11 w/m^2. °c
answer : 44.83 watts
Explanation:
Given data :
surface emissivity ( ε )= 0.95
head ( sphere) diameter( D ) = 0.25 m
Temperature of sphere( T ) = 35° C
Temperature of surrounding ( T∞ ) = 25°C
Temperature of surrounding surface ( Ts ) = 15°C
б = ( 5.67 * 10^-8 )
Determine the total rate of heat loss
First we calculate the surface area of the sphere
As = [tex]\pi D^{2}[/tex]
= [tex]\pi * 0.25^2[/tex] = 0.2 m^2
next we calculate heat loss due to radiation
Qrad = ε * б * As( [tex]T^{4} - T^{4} _{s}[/tex] ) ---- ( 1 )
where ;
ε = 0.95
б = ( 5.67 * 10^-8 )
As = 0.2 m^2
T = 35 + 273 = 308 k
Ts = 15 + 273 = 288 k
input values into equation 1
Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )
= 22.83 watts
Qrad ( heat loss due to radiation ) = 22.83 watts
calculate the heat loss due to convection
Qconv = h* As ( ΔT )
= 11*0.2 ( 35 -25 ) = 22 watts
Hence total rate of heat loss
= 22 + 22.83
= 44.83 watts
Five identical keys are suspended from a balance, which is held horizontally as shown. The two keys on the left are attached to the balance 6 cm from the pivot and the three keys on the right are attached 5 cm from the pivot. What will happen when the person lets go of the balance beam?
Answer:
movement in clockwise direction.
Explanation:
The following parameters or information are given from the question above, they are:
[1]. There are two identical keys, [2]. two out of the five keys are attached to 6cm from the pivot, [3]. three keys out of the five keys on the right are attached 5 cm.
Therefore, considering the moment of force, the two keys on the left = 2 × 6 = 12.
Also, considering the moment of force, the 3 keys on the right = 3 × 5 = 15.
Therefore, we have more weight on the right keys. So, in order to balance the force there must be movement in clockwise direction.
Select the correct answer.
Tires need to be regularly:
A.
Replaced
B.
Rotated
C.
Both A and B
D.
None of the above
Determine the size of memory needed for CD recording of a piece of music, which lasts for 26 minutes, is done with a 20-bit Analog-to-Digital Converter (ADC) in stereo (2 channels), at the rate of 44.1 kSa/s, with the compression factor 6 (allow 10% error margin).
Answer: the size of memory needed for the CD recording is 28.7 MB
Explanation:
so in the case of stereo, the bitrate is;
⇒ 26 × 60 × 44.1 × 10³ × 2
= 137592 × 10³
for 10 bit
⇒ 137592 × 10³ × 10
= 1375920 × 10³ bits
now divide by 8 (convert to bytes)
⇒ (1375920 × 10³) / 8
= 171,990,000 BYTE
divide by 1000 (convert to kilobytes)
= 171,990,000 / 1000
= 171,990 KILOBYTES
now Given that, the compression ratio is 6
so
171,990 / 6
= 28665 KB
we know that. 1 MB = 1000 KB
x MB = 28665 KB
x MB = 28665 / 1000
⇒ 28.665 MB ≈ 28.7 MB
Therefore the size of memory needed for the CD recording is 28.7 MB
In a compression test, a steel test specimen (modulus of elasticity 30 106 lb/in2 ) has a starting height 2.0 in and diameter 1.5 in. The metal yields (0.2% offset) at a load 140,000 lb. At a load of 260,000 lb, the height has been reduced to 1.6 in. Determine (a) yield strength and (b) fl ow curve parameters (strength coeffi cient and strain-hardening exponent). Assume that the cross-sectional area increases uniformly during the test.
Answer:
A) σ_y = 79096 lb/in² = 79.1 ksi
B) strain-hardening exponent = 0.102
(strength coefficient = 137838.78 lb/in²
Explanation:
A) Formula for volume is;
V = πd²h/4
We are given;
height 2.0 in and diameter 1.5 in
Thus;
V = (π × 1.5² × 2)/4
V = 3.53 in³
Area is;
A = πd²/4
A = (π × 1.5²)/4
A = 1.77 in²
Yield strength is gotten from the formula;
σ_y = Force/Area
We are given load = 140,000 lb
Thus;
σ_y = 140000/1.77
σ_y = 79096 lb/in²
B) We are given
modulus of elasticity: E = 30 × 10^(6) lb/in²
Formula for strain is;
ε = σ_y/E
ε = 79096/(30 × 10^(6))
ε = 0.00264
The metal yields (0.2% offset), thus;
strain offsets = 0.00264 + 0.002
strain offsets: ε1 = 0.00464
Thus;
(h_i - h_o)/h_o = 0.00464
(h_i/h_o) - 1 = 0.00464
(h_i/h_o) = 1.00464
h_i = h_o(1.00464)
h_o = 2 in
Thus; h_i = 2(1.00464) = 2.00928 in
Area = Volume/height = 3.53/2.00928 = 1.757 in²
True stress is;
σ = force/area = 140000/1.757
σ1 = 79681.27 lb/in²
At a load of 260,000 lb, the height has been reduced to 1.6 in. Thus;
Area = 3.53/1.6 = 2.206 in²
True stress is;
σ2 = 260000/2.206
σ2 = 117860.38 lb/in²
True strain;
ε2 = In(2/1.6)
ε2 = 0.223
From flow curve;
σ = kεⁿ
Thus;
σ1 = k(ε1)ⁿ
79681.27 = k(0.00464ⁿ) - - - (eq 1)
Also for σ2 = k(ε2)ⁿ;
117860.38 = k(0.223ⁿ) - - - - - (eq 2)
From eq 1,
k = 79681.27/0.00464ⁿ
Putting this for k in eq2 to get;
117860.38 = (0.223ⁿ) × 79681.27/0.00464ⁿ
117860.38/79681.27 = 0.223ⁿ/0.00464ⁿ
Solving for n, we have ≈ 0.102
Thus,K is;
k = 79681.27/0.00464^(0.102)
k = 137838.78 lb/in²
What phenomenon allows water to reach the top of a building?
greywater
venting
water pressure
Owater vapor
Answer:
Option C: water pressure.
Explanation:
Water pressure allows water to reach the top of a building.
An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What are the dimensions of the box so that the total surface area (of all six sides) of the box is minimized
Answer:
[tex]Length =3[/tex] [tex]Height = 2[/tex] and [tex]Width = \frac{3}{2}[/tex]
Explanation:
Given
[tex]Volume = 9m^3[/tex]
Represent the height as h, the length as l and the width as w.
From the question:
[tex]Length = 2 * Width[/tex]
[tex]l = 2w[/tex]
Volume of a box is calculated as:
[tex]V = l*w*h[/tex]
This gives:
[tex]V = 2w *w*h[/tex]
[tex]V = 2w^2h[/tex]
Substitute 9 for V
[tex]9 = 2w^2h[/tex]
Make h the subject:
[tex]h = \frac{9}{2w^2}[/tex]
The surface area is calculated as:
[tex]A = 2(lw + lh + hw)[/tex]
Recall that: [tex]l = 2w[/tex]
[tex]A = 2(2w*w + 2w*h + hw)[/tex]
[tex]A = 2(2w^2 + 2wh + hw)[/tex]
[tex]A = 2(2w^2 + 3wh)[/tex]
[tex]A = 4w^2 + 6wh[/tex]
Recall that: [tex]h = \frac{9}{2w^2}[/tex]
So:
[tex]A = 4w^2 + 6w * \frac{9}{2w^2}[/tex]
[tex]A = 4w^2 + 6* \frac{9}{2w}[/tex]
[tex]A = 4w^2 + \frac{6* 9}{2w}[/tex]
[tex]A = 4w^2 + \frac{3* 9}{w}[/tex]
[tex]A = 4w^2 + \frac{27}{w}[/tex]
To minimize the surface area, we have to differentiate with respect to w
[tex]A' = 8w - 27w^{-2}[/tex]
Set A' to 0
[tex]0 = 8w - 27w^{-2}[/tex]
Add [tex]27w^{-2}[/tex] to both sides
[tex]27w^{-2} = 8w[/tex]
Multiply both sides by [tex]w^2[/tex]
[tex]27w^{-2}*w^2 = 8w*w^2[/tex]
[tex]27 = 8w^3[/tex]
Make [tex]w^3[/tex] the subject
[tex]w^3 = \frac{27}{8}[/tex]
Solve for w
[tex]w = \sqrt[3]{\frac{27}{8}}[/tex]
[tex]w = \frac{3}{2}[/tex]
Recall that : [tex]h = \frac{9}{2w^2}[/tex] and [tex]l = 2w[/tex]
[tex]h = \frac{9}{2 * \frac{3}{2}^2}[/tex]
[tex]h = \frac{9}{2 * \frac{9}{4}}[/tex]
[tex]h = \frac{9}{\frac{9}{2}}[/tex]
[tex]h = 9/\frac{9}{2}[/tex]
[tex]h = 9*\frac{2}{9}[/tex]
[tex]h= 2[/tex]
[tex]l = 2w[/tex]
[tex]l = 2 * \frac{3}{2}[/tex]
[tex]l = 3[/tex]
Hence, the dimension that minimizes the surface area is:
[tex]Length =3[/tex] [tex]Height = 2[/tex] and [tex]Width = \frac{3}{2}[/tex]
How many flip-flop values are complemented in an 8-bit binary ripple counter to reach the next count value after: 0110111 and 01010110?
Answer:
- Four (4) flip-flop values will complemented
- one (1) flip-flop value will complemented
Explanation:
To find how many flip flop number of bits complemented, we just need to figure out what the next count in the sequence is and find how many bits have changed.
taking a look at the a) 00110111
we need to just 1 to the value,
so
00110111 + 0000001 = 00111000
So here, only the first four bits are complemented.
Therefore Four (4) flip-flop values will complemented
Next
b) 01010110
we also add 1 to the value
01010110 + 00000001 = 01010111
only the first bit is complemented.
Therefore one (1) flip-flop value will complemented
eggzOG43199
:) it has to be more than 20 characters so I'm just saying this
Answer:
giberishgiberishgiberishgiberishgiberishgiberishgiberishgiberishgiberish
Explanation:
i have no idea what the question is but do u 4 real need help?
The purpose of pasteurizing milk is to A. Kill pathogens B. Break down milk fat C. Add vitamins and minerals D. Prevent spoilage by sunlight
LOLOLOLOKOLLOLLOLOLOO STRIKER KID THINKS HES SO GOOD LLOLOLOLOLOLOLOLOLOLOOLOLOLOLOLOLOL
Answer:
UUUUUUMMMM do you mean in soccer ????????????????
Explanation:
An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205 m). By how much (in newtons) does her weight change when she goes from sea level to the observatory?
Answer:
[tex]0.845\ \text{N}[/tex]
Explanation:
g = Acceleration due to gravity at sea level = [tex]9.81\ \text{m/s}^2[/tex]
R = Radius of Earth = 6371000 m
h = Altitude of observatory = 4205 m
Change in acceleration due to gravity due to change in altitude is given by
[tex]g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2[/tex]
Weight at sea level
[tex]W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}[/tex]
Weight at the given height
[tex]W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}[/tex]
Change in weight [tex]W_h-W=636.805-637.65=-0.845\ \text{N}[/tex]
Her weight reduces by [tex]0.845\ \text{N}[/tex].
Which of the following terms describes the path from an electrical source to a switch or plug?
transmitter
circuit breaker
raceway
breaker panel
Answer:
transmitter hope thus helped!
Explanation:
Raceway is the answer
"A raceway is an enclosed conduit that forms a physical pathway for electrical wiring."
It is desired to produce and aligned carbon fiber-epoxy matrix composite having a longitudinal tensile strength of 630 MPa. Calculate (a) the critical fiber length, and (b) the volume fraction of fibers necessary if (1) the average fiber diameter is 0.030 mm, (2) the average fiber length is 2.4 mm, (3) the fiber fracture strength is 5100 MPa, (4) the fiber-matrix bond strength is 17 MPa, (5) the matrix stress at fiber failure is 17.5 MPa.
Answer:
The answer is below
Explanation:
Given that:
Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength [tex](\sigma_{cd})=630\ MPa = 630*10^6\ Pa[/tex], Fracture strength
[tex](\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa[/tex]
a) The critical length ([tex]L_c[/tex]) is given by:
[tex]L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm[/tex]
The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.
b) The volume fraction (Vf) is gotten from the formula:
[tex]\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456[/tex]