Hi, can anyone draw me an isometric image of this shape?
A brittle material is subjected to a tensile stress of 1.65 MPa. If the specific surface energy and modulus of elasticity for this material are 0.60 J/m2 and 2.0 GPa, respectively. What is the maximum length of a surface flaw that is possible without fracture
Answer:
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
Explanation:
The given values are,
σ=1.65 MPa
γs=0.60 J/m2
E= 2.0 GPa
The maximum possible length is calculated as:
[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM signal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier
Answer:
"192.3 watt" is the right answer.
Explanation:
Given:
Efficient amplifier,
= 65%
or,
= 0.65
Power,
[tex]P_c=250 \ watt[/tex]
As we know,
⇒ [tex]P_t=P_c(1+\frac{\mu^2}{2} )[/tex]
By putting the values, we get
[tex]=P_c(1+\frac{1}{2} )[/tex]
[tex]=1.5 \ P_c[/tex]
Now,
⇒ [tex]P_i=(P_t-P_c)[/tex]
[tex]=1.5 \ P_c-P_c[/tex]
[tex]=\frac{P_c}{2}[/tex]
DC input (0.65) will be equal to "[tex](\frac{P_c}{2} )[/tex]".
hence,
The DC input power will be:
= [tex]\frac{250}{2}\times \frac{1}{0.65}[/tex]
= [tex]\frac{125}{0.65}[/tex]
= [tex]192.3 \ watt[/tex]
4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The ambient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin
Answer:
a) 84.034°C
b) 92.56°C
c) ≈ 88 watts
Explanation:
Thickness of aluminum alloy fin = 12 mm
width = 10 mm
length = 50 mm
Ambient air temperature = 22°C
Temperature of aluminum alloy is maintained at 120°C
a) Determine temperature at end of fin
m = √ hp/Ka
= √( 140*2 ) / ( 12 * 10^-3 * 55 )
= √ 280 / 0.66 = 20.60
Attached below is the remaining answers
What is the per capita GDP of China? Be sure to indicate the calendar year that this information represents.
The per capita GDP of China in the Calendar year 2021 was found to be around 12,359 U.S. dollars.
What is GDP?GDP termed Gross Domestic Product, has been evaluated with the value producing the economy of the region with the values added with the used products formed to be the less of the economy produced. It has been termed as the measure of the income of a region and not the wealth.
The per capita GDP has been the total income earned by a person in a region during a specified period of time. The calculation has been made by dividing the total gross income of the region by the total population.
China has been the world's most populous country in the East Asian region. It has been found that the per capita GDP of China is low because of its large population. In the calendar year 2021, the per capita GDP of China was 12,359 U.S. dollars.
Learn more about the GDP, here:
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How to Cancel prescription
If you deposit $ 1000 per month into an investment account that pays interest at a rate of 9% per year compounded quarterly.how much will be in your account at the end of 5 years ?assume no interpèriod compounding
Answer:
5,465.4165939453
Explanation:
formula
A=P(1+r/n)^n(t)
p=1000
r=0.09
n=4
t=5
: Một nền kinh tế có cấu trúc như sau:
C = 80 + 0,8(Y - T); T = 100 ;
I = 130; G = 120;
MSr = MS/CPI = 200;
MD = 0,2Y – 10i
Yêu cầu:
1. Xác định thu nhập và lãi suất cân bằng?
2. Muốn sản lượng cân bằng tăng 500 thì chính phủ cần thay đổi thuế như thế nào?
3. Liệu mục tiêu ở câu 2 có thể đạt đựơc bằng chính sách tiền tệ hay không? Tại sao?
Answer:
Haha I'm a great guy but my friend has been in a day of the day and a lot to be able and I'm happy holi and a lot to the world of the day and day to
What is the key objective of data analysis
Answer: The process of data analysis uses analytical and logical reasoning to gain information from the data. The main purpose of data analysis is to find meaning in data so that the derived knowledge can be used to make informed decisions.
A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft above that of the first. the total length of pipes required is 1000 ft. All pipes are 12 in. in diameter and are made of galvanized iron with relative roughness equal to 0.0005 (you may assume fully-rough flow). the pump performance curves suggest that the H-Q relationship is of the form: H_pump=665-0.051Q^2 (Q in ft) the expected flow rate the brake horsepower required to drive the pump (assume an efficiency of 78%). the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).
Answer:
a) the expected flow rate is 31.4 ft³/s
b) the required brake horsepower is 2808.4 bhp
c) the location of pump inlet to avoid cavitation is -8.4 ft
Explanation:
Given the data in the question;
free surface elevation = 200 ft
total length of pipe required = 1000 ft
diameter = 12 inch
Iron with relative roughness ( k/D ) = 0.0005
H[tex]_{pump[/tex] = 665-0.051Q² [Qinft ]
a) the expected flow rate
given that;
k/D = 0.0005
k/2R = 0.0005
R/k = 1000
now, we determine the friction factor;
1/√f = 2log₁₀( R/k ) + 1.74
we substitute
1/√f = 2log₁₀( 1000 ) + 1.74
1/√f = 6 + 1.74
1/√f = 7.74
√f = 1/7.74
√f = 0.1291989
f = (0.1291989)²
f = 0.01669
Now, Using Bernoulli theorem between two reservoirs;
(p/ρq)₁ + (v²/2g)₁ + z₁ + H[tex]_p[/tex] = (p/ρq)₂ + (v²/2g)₂ + z₂ + h[tex]_L[/tex]
so
0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²
665-0.051Q² = 200 + flQ²/2gdA²
665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )
665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]
665 - 200 - 0.051Q² = 0.420138Q²
665 - 200 = 0.420138Q² + 0.051Q²
465 = 0.471138Q²
Q² = 465 / 0.471138
Q² = 986.97196
Q = √986.97196
Q = 31.4 ft³/s
Therefore, the expected flow rate is 31.4 ft³/s
b) the brake horsepower required to drive the pump (assume an efficiency of 78%).
we know that;
P = ρgH[tex]_p[/tex]Q / η
where; H[tex]_p[/tex] = 665 - 0.051(986.97196) = 614.7
we substitute;
P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )
P = 1204804.6236 / 429
P = 2808.4 bhp
Therefore, the required brake horsepower is 2808.4 bhp
c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).
NPSH = ([tex]P_{atom[/tex] / ρg) - h[tex]_s[/tex] - ( P[tex]_v[/tex] / ρg )
we substitute
25 = ( 2116 / 62.42 ) - h[tex]_s[/tex] - ( 30 / 62.42 )
h[tex]_s[/tex] = 8.4 ft
Therefore, the location of pump inlet to avoid cavitation is -8.4 ft
