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ludmilkaskok [199]
3 years ago
15

Technician A says that when using an impact wrench to remove a bolt from the front of an engine's crankshaft, the crankshaft mus

t be held to keep it from turning. Technician B says that when using an impact wrench to remove lug nuts on a raised vehicle, the brakes must be applied. Who is right?
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Engineering
1 answer:
Lady bird [3.3K]3 years ago
3 0

Answer:

d. Neither A nor B

Explanation:

Mostly for exhaust bolts we use impact wrenches. Impact wrenches are very useful on flywheels because the crank shaft needs to be fixed from rotating. Also impact wrenches are useful on cars for loosening wheel nuts.  The wheels need not to be restrained or the brakes need not to be applied to fiz wheels in position.

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A permanent magnet DC motor has an armature resistance of 0.5 Ω and when a voltage of 120 V is applied to the motor it reaches a
BaLLatris [955]

Answer:

12 N-m

Explanation:

The dc motor is operating at 24 V that is its terminal voltage V =24 V

Armature resistance  = 0.2 ohm

No load speed = 240 radian /sec

For motor we know that  as the motor is on no load so  so

Power developed in the motor

Now we know that power = torque× angular speed

So

3 0
4 years ago
To practice my skills, Jesse asked me to create an imaginary project with at least 12 tasks, which include dependent, multiple p
Sati [7]

Answer: see attachment

Explanation:

8 0
4 years ago
A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici
Alika [10]

Answer:

The maximum length of a surface flaw is 8.24 μm

8.24 μm

Explanation:

Given that:

The modulus of elasticity E = 69 GPa

The specific surface energy \delta_s = 0.3 J/m²

The length of the surface flaw "a" = ??

From the theory of the brittle fracture;

\sigma _c = \bigg (  \dfrac{2E \delta_s}{\pi a}  \bigg )^{1/2}

Making a the subject of the formula; we have:

a = \bigg (  \dfrac{2 \times E \times \delta_s}{\pi \sigma _c ^2}  \bigg )

a= \bigg (  \dfrac{2 \times 69*10^9 \times 0.3}{\pi (40*10^6)^2}  \bigg )

a = 8.24 × 10⁻⁶ m

a = 8.24 μm

Thus; the maximum length of a surface flaw is 8.24 μm

3 0
3 years ago
Calculate the unit length of a roof using common rafters if the unit run is 7 inches. Round your answer to two decimal places.
elena-14-01-66 [18.8K]

Answer:

Explanation:

https://www.pole-barn.info/roof-rafter-calculations.html

4 0
3 years ago
A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u
Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}

where,

\gamma= Specific Heat ratio of air

M = Mach number upstream

We know that \gamma _{air}=1.4

Applying values we get

\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)

Similarly the temperature downstream is obtained by the relation

\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

Applying values we get

\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

3 0
4 years ago
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