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tensa zangetsu [6.8K]
3 years ago
5

- You have a bin wrench turning a 1/2 13 UNC bolt. You overcome 1200 lbs of resistance when you

Engineering
1 answer:
andrew11 [14]3 years ago
6 0

Fr

is my guess but yeah

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(Blank) welding involves manual welding with equipment anomalously controls one or more of the windy conditions while (blank) we
shepuryov [24]

Answer:

D.

Explanation:

In automated welding, defined as “welding with equipment that requires only occasional or no observation of the weld, and no manual adjustment of the equipment controls,” the welder's involvement is limited to activating the machine to initiate the welding cycle and observing the weld on an intermittent basis.

8 0
3 years ago
The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:
Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1

Here;

G_{s} is specific gravity of soil solids

y_{w} is unit weight of water = 998 kg/m^3

w is the moisture content = 0.17

e is the void ratio

y is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

w*G_{s} = S*e  ..... Eq2

S is saturation rate

Substitute Eq 2 into Eq 1

y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w}  }{1+e}

14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233

Specific gravity of soil solids

G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765

Saturated Unit Weight

y_{s} = \frac{(G_{s} + e)*y_{w}  }{1+e} \\=\frac{(4.878811765 + 1.38233)*998  }{1+1.38233}\\\\= 2622.902571 N/m^3

7 0
3 years ago
Please solve part two
Burka [1]

Answer:

Wat part 2

Explanation:

7 0
3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2
IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length l for the deformation, we use the following relation;

l = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

l = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

l = 3161.025289 / 7240

l = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0
3 years ago
There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter
weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

                                        u_e = 9.30777 * 10^-8  J/m^3

5 0
3 years ago
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