Technician A says that a continuously variable transmission is an automatic transmission that does not shift gears. Technician B says that a continuously variable transmission has a maximum of six distinct gear ranges. Which technician is correct?

Complete question:

A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?

a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.

b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.

c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.

d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.

Answer:

A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.

Explanation:

we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.

thank you!

Answer:

5.16 M

Explanation:

Loads ; 1800N, 2100N, 2275N, 2200N

safety factor = 5

diameter of each column = 50 cm = 0.5 m

Elastic modulus = 10 MPa

Calculate the max height of structure

moment of inertia for a circular section ( I ) = πd^4 / 64

lets represent the required maximum height of the column as L

Applying Euler column theory

The bucker load of the column =  ( attached below )

attached below is the remaining solution

Explanation:

Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.

This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.

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Example

Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...

  mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)

  mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs

  mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0

You will note that the matrix of equation coefficients is symmetric.

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In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.

Answer:

can you plz write in English language so we can give you answer

Answer:

The correct answer is "57%".

Explanation:

Given:

Temperature:

[tex]T_a = 298 \ K[/tex]

[tex]T_3 = 2000 \ K[/tex]

Compression ratio (r),

= [tex]8.2[/tex]

For Otto cycle, the thermal efficiency will be:

⇒ [tex]\eta =1-\frac{1}{(r)^{\nu-1}}[/tex]

By substituting the values, we get

       [tex]=1-\frac{1}{(8.2)^{1.4-1}}[/tex]

       [tex]=1-\frac{1}{(8.2)^{0.4}}[/tex]

       [tex]=57[/tex] (%)  

Answer:

65.61%

Explanation:

we have the following information to answer this question

diameter of the solid steel shaft = 40 mm

outer diametr of the hollow shaft = 40mm

inner diametr pf the hollow shaft = 36mm

[tex]percentage reduction in torque transmission = \frac{Tsolid-Thollow}{Tsolid} *100[/tex]

= (40³ - (40⁴-36⁴)/40)/40³ * 100

= (40³ - 22009.6)/40³ * 100

= 41990.4/64000 * 100

= 0.6561 x 100

= 65.61%

percentage reduction in torque transmission = 65.61%

Answer:

10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

The question is incomplete. The complete question is :

The base of an aluminum block, which is fixed in place, measures 90 cm by 90 cm, and the height of the block is 60 cm. A force, applied to the upper face and parallel to it, produces a shear strain of 0.0060. The shear modulus of aluminum is [tex]3.0 \times 10^{10} \ Pa[/tex] . What is the displacement of the upper face in the direction of the applied force?

Solution :

The relation between shear modulus, shear stress and strain,

[tex]$\text{Shear modulus, S =} \frac{\text{Shear stress}}{\text{shear strain}}$[/tex]

Shear stress = shear modulus (S) x shear strain

                     [tex]$=3 \times 10^{10} \times 0.0060$[/tex]

                     [tex]$=1.80 \times 10^8$[/tex] Pa

                     [tex]$=180 \times 10^6$[/tex] Pa

                     [tex]$=180 \ MPa$[/tex]

The length represents the distance between the fixed in place portion and where the force is being applied.

Therefore,

[tex]$\text{Displacement} = \text{shear strain} \times \text{length}$[/tex]

                     = 0.006 x 60 cm

                      = 0.360 cm

                      = 3.6 mm

Thus, the displacement of the upper face is 3.6 mm in the direction of the applied force.

spanish

Explanation:

the above question is written in spanish

Complete Question

Complete Question is attached below

Answer:

[tex]P=1124.2ibf[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=1.2in[/tex]

Allowable Normal stress [tex]\sigma=27.5ksi=27.5 * 10^3 psi[/tex]

Generally the equation for Bending Stress is mathematically given by

[tex]\phi= \frac{32M}{ \pi d^3}[/tex]

[tex]\phi= {32 * 4 P}{\pi * 1.2^3}[/tex]

[tex]\phi=23.58 psi[/tex]

Generally the equation for Direct Normal Stress is mathematically given by

[tex]\sigma'=\frac{4P}{ \phi * 1.2^2}[/tex]  

[tex]\sigma'= 0.88P psi[/tex]

Therefore

Total Normal stress

[tex]\sigma_T=23.58 + 0.88[/tex]

[tex]\sigma_T=24.46P[/tex]

Generally the equation for Allowable Stress is mathematically given by

[tex]\sigma=\sigma_T P[/tex]

[tex]P=\frac{\sigma}{\sigma_T}[/tex]

[tex]P=\frac{27.5 * 10^3}{24.46P}[/tex]

[tex]P=1124.2ibf[/tex]

Answer:

Data Rate Signaling

Answer:

No it is not permitted

Explanation:

It is not permitted  because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.

The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.

Answer:

[tex]T_2=315.69k[/tex]

Explanation:

Initial Temperature [tex]T_1=500K[/tex]

Initial Pressure [tex]P_1=1000kPa[/tex]

Final Pressure [tex]P_2=200kPa[/tex]

Generally the gas equation is mathematically given by

[tex]\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}[/tex]

Where

n for [tex]CO=1.4[/tex]

Therefore

[tex]\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}[/tex]

[tex]T_2=315.69k[/tex]

Answer: The tracer lathe is a roughing operation for the output shaft on rear wheel drive transmissions.

Explanation:

Answer:

Explanation:

From the given information:

weight of fiber [tex]w_f[/tex] = 3.0 g

weight of composite specimen [tex]w_c[/tex] = 4.0 g

specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g

specific gravity of fiber [tex]S_f[/tex] = 2.4

specific gravity of matrix [tex]S_m[/tex] = 1.3

The weight of the matrix = weight of the composite - the weight of fiber

⇒ (4.0 - 3.0) g

= 1.0 g

The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]

[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]

[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]

[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]

[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]

The experimental density [tex]\rho _{ce}[/tex] is determined  by using the equation:

[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]

[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]

[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]

The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]

[tex]= \dfrac{1.980-1.620}{1.980}[/tex]

= 0.1818

Answer:

[tex]D=0.41m[/tex]

Explanation:

From the question we are told that:

Discharge rate [tex]V_r=0.35 m3/s[/tex]

Distance [tex]d=4km[/tex]

Elevation of the pumping station [tex]h_p= 140 m[/tex]

Elevation of the Exit point [tex]h_e= 150 m[/tex]

Generally the Steady Flow Energy Equation SFEE is mathematically given by

[tex]h_p=h_e+h[/tex]

With

[tex]P_1-P_2[/tex]

And

[tex]V_1=V-2[/tex]

Therefore

[tex]h=140-150[/tex]

[tex]h=10[/tex]

Generally h is give as

[tex]h=\frac{0.5LV^2}{2gD}[/tex]

[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

Therefore

[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]

[tex]D=0.41m[/tex]

Answer:

click here and I'm so happy I have been doing this weekend I can see the same way you could come back in a bit and then we are all of you to know you do for a bit of time with me or

Explanation:

there are three stages of heat treatment

1.hit the metal slowly to ensure that the metal maintains a uniform temperature

2.soak or hold the metal at a specific temperature for a alloted period of time

3.cool the metal to room temperature

Answer:

hey. its a big question. solved from *c hegg

Explanation:

Answer:

An OTG or On The Go adapter (sometimes called an OTG cable, or OTG connector) allows you to connect a full sized USB flash drive or USB A cable to your phone or tablet through the Micro USB or USB-C charging port

Explanation:

pls mark brainliest

Answer:

Can you make friend with me ?

Answer:

348 g

Explanation:

The heat gained by the ice equals the heat lost by the hot tea.

Also, since we require a very, very small amount of ice and some water, to still have some ice, the temperature of the hot tea has to decrease to the melting point of ice which is 0°C. So, the final temperature of the mixture is 0 °C.

So, mc(T - T') + mL = -MC(T - T") where m = mass of ice, c =  specific heat of ice = 2090 J/kgK., T = final temperature of mixture = 0 °C, T' = initial temperature of ice = -12 °C, L = heat of fusion of H2O = 3.33 × 10⁵ J/kg, M = mass of hot tea = ρV where ρ = density of tea = 1.00 g/mL and V = volume of hot tea = 350 mL. So, M = ρV = 1.00 g/mL × 350 mL = 350 g = 0.350 kg, C = specific heat of tea = 4190 J/kgK and T" = initial temperature of tea = 85 °C.

Making m subject of the formula, we have

mc(T - T') + mL = -MC(T - T")

m[c(T - T') + L] = -MC(T - T")  

m = -MC(T - T")/[c(T - T') + L]

substituting the values of the variables into the equation, we have

m = -MC(T - T")/[c(T - T') + L]

m = -0.350 kg × 4190 J/kgK(0 °C - 85 °C.)/[2090 J/kgK(0 °C. - (-12 °C)) + 3.33 × 10⁵ J/kg]

m = -1466.5 J/kK(- 85 °C)/[2090 J/kgK(0 °C + 12 °C)) + 3.33 × 10⁵ J/kg]

m = 124652.5 J/[2090 J/kgK(12 °C) + 3.33 × 10⁵ J/kg]

m = 124652.5 J/[25080 J/kg + 3.33 × 10⁵ J/kg]

m = 124652.5 J/358080 J/kg

m = 0.3481 kg

m = 348.1 g

m ≅ 348 g

So, the minimum amount of ice to be added is 348 g.

Answer: The options related to your question is missing attached below is the missing option

a.) The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface.

b.) The insulation must be arranged in order of decreasing k starting with the highest k material at the wall surface.

c.) The insulation can be arranged in any order.

d.) It is impossible to decide because the temperatures are not specified and the order will depend on the ∆T.

e) The insulation with the lowest k should be nearest the wall, and the order of the other two is irrelevant.

answer:

The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface. ( A )

Explanation:

The correct statement is ; The insulation must be arranged in order of increasing k starting with the lowest k material at the wall surface.

Because the insulation has to be arranged following the order of it's increasing thermal conductivity ( K-value )