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4 years ago

Which graph represents the fastest runner

Engineering

2 answers:

Arada [10]4 years ago

5 0

Answer:

Graph c represents the fastest runner

Basile [38]4 years ago

4 0

Answer: B

Explanation:

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ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in

Mama L [17]

Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

$d_{p} =mN_{p}$

Where

m = module = 5

Np = number of teeth of pinion = 26

$d_{p} =5*26=130mm$ = 0.13 m

The pitch line velocity is equal to:

$V_{t} =\frac{d_{p}*2*\pi *w_{p} }{120}$

Where

wp = speed of the pinion = 1800 rpm

$V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s$

The factor B is equal to:

$B=\frac{(12-Q_{v})^{2/3} }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396$

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

$K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832$

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

$P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}$

The bending stress is equal to:

$\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}} \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa$

4 0

3 years ago

A vertical cylinder (Fig. P3.227) has a 61.18-kg piston locked with a pin, trapping 10 L of R-410a at 10◦C with 90% quality insi

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4 0

3 years ago

A small vehicle is powered by a pulsejet. The available net thrust is 6000 N and the traveling speed is 200 km/hr. The gases lea

kumpel [21]

Answer:

a) The mass flow rate is 19.71 kg/s

b) The inlet area is 0.41 m²

c) The thrust power is 333.31 kW

d) The propulsive efficiency is 26.7%

Explanation:

Please look at the solution in the attached Word file.

Download docx

4 0

3 years ago

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

miss Akunina [59]

Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

b) Temperature of interface: $T_{interface}=35.6\°C$

Heat flux through skin: $\frac{Q}{A}=2100W/m^2$

Explanation:

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side