Which of the following vehicles has no emissions?

A strain gauge with a 5 mm gauge length gives a displacement reading of 1.25 um. Calculate the stress value given by this displa

KengaRu [80]

Answer:

stress = 50MPa

Explanation:

given data:

Length of strain guage is 5mm

displacement $\delta = 1.25 \mu m =\frac{1.25}{1000} = 0.00125 mm$

stress due to displacement in structural steel can be determined by using following relation

$E =\frac{stress}{strain}$

$stress = E \times strain$

where E is young's modulus of elasticity

E for steel is 200 GPa

$stress = 200\times 10^3 *\frac{1.25*10^{-3}}{5}$

stress = 50MPa

7 0

3 years ago

While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo

IgorC [24]

Answer:

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

$z=12x\frac{1}{6}+2x\frac{1}{2}+3=6$

z = 6 atoms per unit cell

M = 91.224 g/mol

N = $6.023x10^{23} atoms/mol$

d = $6.51g/cm^{3}$

$V_{c}=\frac{zxM}{dxN}$

$V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }$

$V_{c}=1.396x10^{-22} cm^{3} /unit.cell$

$V_{c}=1.396x10^{-28} m^{3} /unit.cell$

8 0

3 years ago

Select the correct answer.

andre [41]

Answer:

A. energy transformations

Explanation:

8 0

2 years ago

For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5

bonufazy [111]

Answer:

The total force resisted by the flange bolts is 163.98 N

Explanation:

Solution

The first step is to find the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ = diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of the pipe at the inlet section

A₂ = π /4 D₂²

D₂ = diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265 * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0

2 years ago

The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

5 0

2 years ago