O que é necessário se conhecer para a determinação de uma força concentrada em um ponto?

An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tens

Natalka [10]

Answer:

It will results in mechanical hardening.

5 0

3 years ago

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Explain the prosses of welding

klio [65]

Answer:

Welding is a fabrication process whereby two or more parts are fused together by means of heat, pressure or both forming a join as the parts cool. Welding is usually used on metals and thermoplastics but can also be used on wood.

Explanation:

7 0

3 years ago

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Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha

Romashka-Z-Leto [24]

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

8 0

3 years ago

Name some technical skills that are suitable for school leavers .

Natalka [10]

Answer:

Welding, carpentry, masonry, construction worker, barber

Explanation:

7 0

3 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$