8. Removing damage by tapping it out with a hammerand dolly is often referred to as
(A) contouring
(B) roughing out
(C) forming
(D) tapping out

Answers

Answer 1
Answer:

Answer:

tapping out I think sorry if I'm wrong


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The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms in any proportion without altering the crystal structure except by expanding it slightly: this is an example of a binary solid solution series. For different compositions, the lines in the powder diffraction patterns are in slightly different positions, because of the cell expansion, but the overall pattern remains basically the same. The spacing of the lattice planes varies linearly with composition, and this can be used in a rapid and non- destructive method of analysis. a. The (062) reflection from olivine is strong and well resolved from other lines. Calculate d062 for an olivine that displays its (062) reflection at a Bragg angle of 37.21° (i.e., a diffraction angle of 74.42°) when x-rays with a wavelength of 0.1790 nm are used. b. The d062 spacing as measured accurately for synthetic materials is 0.14774 nm for Mg2SiO4 and 0.15153 nm for Fe2SiO4. What would be the approximate composition, expressed in mol.% Mg2SiO4, of an olivine material for which do62 has the value obtained in part 2.1 above?
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A refrigeration system was checked for leaks. The system temperature and surroundings were 75°F when the system was charged with nitrogen to 100 psig. The temperature then dropped to 50°F. What should the pressure be if no nitrogen has escaped?A) 9 psig
B) 94 psig
C) 100 psig
D) 90 psig

Answers

The temperature and pressure of an ideal gas are directly proportional

The pressure of the system should be in the range B) 94 psig

The given refrigerator parameters are;

The temperature of the system and the surrounding, T₁ = 75 °F = 237.0389 K

The pressure to which the system was charged with nitrogen, P₁ = 100 psig

The temperature to which the system dropped, T₂ = 50 °F = 283.15 K

The required parameter;

The pressure, P₂, of the system at 50°F

Method:

The relationship between pressure and temperature is given by Gay-Lussac's law as follows;

At constant volume, the pressure of a given mass of gas is directly proportional to its temperature in Kelvin

Mathematically, we have;

(P_1)/(T_1) = \mathbf{(P_2)/(T_2)}

Plugging in the values of the variables gives;

\mathbf{(100 \ psig)/(297.0389)} = (P_2)/(283.15)

Therefore;

P_2 = \mathbf{283.15 \, ^(\circ)F * (100 \ psig)/(297.0389\ ^(\circ)F) \approx 95.3 \, ^(\circ)F}

The closest option to the above pressure is option B) 94 psig

Learn more about Gay-Lussac's law here;

brainly.com/question/16302261

B 94 psig because the temperature dropped too

What regulations is OSHA cover under what act

Answers

The law requires employers to provide their employees with working conditions that are free of known dangers. The OSH Act created the Occupational Safety and Health Administration (OSHA), which sets and enforces protective workplace safety and health standards.

A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 10. The transmitted tangential load is 22 kN. Conditions are such that Km = 1.7. The teeth are standard 20-degree, full-depth. The module is 5 and the face width 62 mm. Determine the bending stress when the mesh is at the highest point of single tooth contact.

Answers

Answer:

The bending stress of the face tooth is  \sigma _(bg) = 502.82 MPa

Explanation:

From the question we are told that

        The number of tooth of the pinion is  N_t = 26 \ tooth

         The velocity of rotation is given as \omega_p = 1800 rpm

         The number of tooth is of the gear is  N_g = 55 \ tooth

        The quality level is Q_r = 10

          The transmitted tangential load is F_T = 22\ kN = 22 KN * (1000N)/(1KN) = 22*10^3 N

                                                                    k_m = 1.7

        The angle of the teeth is  \theta_t = 20^o

         The module is  M= 5

         The face width is W_f = 62mm

The diameter of the pinion is mathematically represented as

                d_p = M * N_t

Substituting the values

                d_p = 5 *26

                    = 130 mm = (130)/(1000) = 0.130m

The pitch line velocity is mathematically represented as

                     V_t = (d_p )/(2) (2 \pi \omega_p)/(60)

Substituting values

                          = (0.130)/(2) * (2 * 3.142 * 1800 )/(60)

                          = 12.25\  m/s

Generally the dynamic factor is mathematically represented as

                      K_v = [(A)/(A +√(200V_t) ) ]^B

Now B is a constant that is mathematically represented as

                B = ((12 -Q_r )^(2/3))/(4)

substituting values

                  = ((12- 10 )^(2/3))/(4)

                  =0.3968

A is also a constant that is mathematically represented as

              A = 50 + 56(1 -B)

Substituting values

             = 50 +56 (1- 0.3968)

             = 83.779

Substituting these value into the equation for dynamic factor we have

           K_v = [(83.779)/(83.779 + √(200 * 12.25) ) ]^(0.3968)

                = 0.831

The geometric bending factor for a 20° profile from table

"AGMA Bending Geometry Factor J for 20°, Full -Depth Teeth with HPSTC Loading , Table 2-9"                

That corresponds to 55 tooth gear meshing with 26 pinion is

                   J_g = 0.41

the diameter pitch can be mathematically represented as

              p_d = (1)/(M)

Substituting values

            p_d  = (1)/(5)

                =0.2mm^(-1)

The mathematically representation for gear tooth bending stress in the teeth face is as follows

          \sigma_(bg) = (F_T \cdot p_d )/(W_f * J_g)(K_a K_(dt) )/(K_v) K_s K_B K_t ----(1)

Where W_t is the tangential load

            W_f is the face width

            K_a is the application factor  this is obtained from table "Application Factors, Table 12-17 " and the value  is  K_a  = 1

            K_(dt) is the load distributed factor

            K_s is the size factor

             K_B is the rim thickness factor which is obtained for M which has a value  1

           K_t is the idler

Substituting values into equation 1

     \sigma_(bg) = (22*10^3 *0.2)/(62 * 0.41) * (1 * 1.7 )/(0.831)  * 1 *1 *1.42

            = 502.82  N/mm^2

            = 502.82 * 1000 * (N)/(m^2)

           = 502.82 MPa

           

           

       

 

               

                 

The hot water needs of an office are met by heating tab water by a heat pump from 16 C to 50 C at an average rate of 0.2 kg/min. If the COP of this heat pump is 2.8, the required power input is: (a) 1.33 kW (d) 10.2 kW (b) 0.17 kW (c) 0.041 kW

Answers

Answer:

option B

Explanation:

given,

heating tap water from 16° C to 50° C

at the average rate of 0.2 kg/min

the COP of this heat pump is 2.8

power output = ?

COP = (Q_H)/(W_(in))\nW_(in) = (Q_H)/(COP)\nW_(in) = ((0.2)/(60)* 4.18* (50-16))/(2.8)\nW_(in) = 0.169

the required power input is 0.169 kW or 0.17 kW

hence, the correct answer is option B

Be a qt today lolololololol

Answers

Answer:

you too

Explanation:

thank you for the free points

Benzene vapor at 480°C is cooled and converted to a liquid at 25°C in a continuous condenser. The condensate is drained into 1.75-m3 drums, each of which takes 2.0 minutes to fill. Calculate the rate (kW) at which heat is transferred from the benzene in the condenser.

Answers

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.