Hi, can anyone draw me an isometric image of this shape?
Question
А
Particle of 2kg mass is being pulled across a smooth horizontal
surface by a horizontal force. The force does 24 Joule of work in
increasing
the particle's
velocity from 5m/s
to v m/s. calculate
the value of v and the position of particle
after 15s
Answer:
udhddhdiejebdidjebdhdidh
: 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
Type the correct answer in each box. Spell all words correctly. According to the priority matrix, which tasks should an entrepreneur complete first? According to the priority matrix, entrepreneurs should first complete tasks that are blank and important.
Answer:
Development of creative and develop ideas
Explanation:
First task as an entrepreneur is to be creative and develop ideas. The person must design the product based on which he will develop the business strategy.
The remaining activities such as marketing, fund raising, recruitment etc. comes at a later stage.
Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2. Derive the unit hydrograph by the inverse procedure. Assume a constant baseflow of 550 cfs.
Hour Day 1 Day 2 Day 3 Day 4
Midnight 550 5,000 19,000 550
6 am 600 4,000 1400
Noon 9000 3000 1000
6 pm 6600 2500 750
Answer:
33.56 ft^3/sec.in
Explanation:
Duration = 6 hours
drainage area = 185 mi^2
constant baseflow = 550 cfs
Derive the unit hydrograph using the inverse procedure
first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below
Vdrh = sum of drh * duration
= 29700 * 6 hours ( 216000 secs )
= 641,520,000 ft^3.
next step : Calculate the volume of runoff in equivalent depth
Vdrh / Area = 641,520,000 / 185 mi^2
= 1.49 in
Finally derive the unit hydrograph
Unit of hydrograph = drh / volume of runoff in equivalent depth
= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in
Calculate the biaxial stresses σ1 and σ2 for the biaxial stress case, where ε1 = .0020 and ε2 = –.0010 are determined experimentally on an aluminum member of elastic constants, E = 71 GPa and v = 0.35. Also, determine the value for the maximum shear stress.
Answer:
i) σ1 = 133.5 MPa
σ2 = -2427 MPa
ii) 78.89 MPa
Explanation:
Given data:
ε1 = 0.0020 and ε2 = –0.0010
E = 71 GPa
v = 0.35
i) Determine the biaxial stresses σ1 and σ2 using the relations below
ε1 = σ1 / E - v (σ2 / E) -----( 1 )
ε2 = σ2 / E - v (σ1 / E) -------( 2 )
resolving equations 1 and 2
σ1 = E / 1 - v^2 { ε1 + vε2 } ---- ( 3 )
σ2 = E / 1 - v^2 { ε2 + vε1 } ----- ( 4 )
input the given data into equation 3 and equation 4
σ1 = 133.5 MPa
σ2 = -2427 MPa
ii) Calculate the value of the maximum shear stress ( Zmax )
Zmax = ( σ1 - σ2 ) / 2
= 133.5 - ( - 2427 ) / 2
= 78.89 MPa
The number of pulses per second from IGBTs is referred to as
Determine the voltage which must be applied to a 1 k 2 resistor in order that a current of
10 mA may flow.
Answer:
The correct solution is "20 volt".
Explanation:
Given that:
Current,
I = 10 mA
or,
= [tex]10\times 10^{-3} \ A[/tex]
Resistance,
R = 2 K ohm
or,
= [tex]1\times 10^3 \ ohm[/tex]
Now,
The voltage will be:
⇒ [tex]V=IR[/tex]
By putting the values, we get
[tex]=10\times 10^{-3}\times 2\times 10^{3}[/tex]
[tex]= 20 \ volt[/tex]
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]
