Ask question

Ask question

All categories

3 years ago

15

When removing the balance shaft assembly: Technician A inspects the bearings for unusual wear or damage. Technician B smoothens

a damaged camshaft or balance shaft journal by lightly sanding it. Who is correct

1 answer:

liraira [26]3 years ago

4 0

Answer:

Technicians A

Explanation:

The technician A would also tap the the teeth with a blunt end to break it loose. He also remove the underlying O ring in the front case groove. Technicians A also remove the driven gear bolt that secures the oil pump driven gear to the left balance shaft.

You might be interested in

Engineering uses scientific thinking to solve real-world problems.<br> True<br> False

erik [133]

True, In the STEM field engineers use and think in terms of sciences like physics and math.

8 0

3 years ago

Help me pls with tools

irina1246 [14]

Answer:

How?

Explanation:

3 0

4 years ago

Read 2 more answers

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:

a)

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

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

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

$k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}$

$k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}$

$k_{skin}=1.5W/mK$

b)

$k_{insulation}=\frac{k_{skin}}{2}$

$k_{insulation}=\frac{1.5}{2}$

$k_{insulation}=0.75W/mK$

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

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

$\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}$

$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

$1875T_{interface}=66750$

$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

3 0

3 years ago

If car 3 in the graph above continues to decelerate at 1.2 mph/s, how fast would it be going at t=20 s?

vladimir2022 [97]

Answer:

3sa

Explanation:

asdfasdfas

8 0

3 years ago

Please answer quickly!!!

Olenka [21]

Answer:

c

Explanation:

nice profile picture by the way

4 0

3 years ago