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3 years ago

Why does my benthic gear say to bring this equpiment to an item upgrade specialist wow

Engineering

1 answer:

vivado [14]3 years ago

4 0

Please mark me Brainliest ik I didn’t help but……

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You are an engineer at company XYZ, and you are dealing with the need to determine the maximum load you can apply to a set of bo

qwelly [4]

Answer:

$F_i = 28.8 kN$

Alternating and mean forces on the bolt are 0.85 kN and 29.65 kN respectively.

Mean and alternating stresses in the bulk of the bolt are 378 MPa and 10 MPa.

Safety factor = 5.5

Explanation:

Basic Dimension of Isometric Screw thread" is considered for analysis.

At Nominal diameter, d = 12 mm , $A_t = 84.3mm^{2}$ , $F_i = 0.9A_tS_p$ , $S_p = 380 MPa$

Rolled threads; $K_f = 2.2$

8 0

2 years ago

Question Completion Status:

Mashutka [201]

Answer: c. Centre of pressure

Explanation:

Pressure is applied on a surface when a force is exerted on a particular point on that surface by another object when the two come into contact with each other.

The point where the pressure is applied is known as the centre of the pressure with the force then spreading out from this point much like an epicentre in an earthquake.

6 0

2 years ago

A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of

andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ $SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}$

By putting the value, we get

⇒ $=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}$

⇒ $=3.7258 \ l$

hence,

The heat loss will be:

⇒ $Q=SF\times K(T_2-T_1)$

$=3.7258\times 1\times 1.8\times (500-300)$

$=3.7258\times 1.8\times (200)$

$=1341.288 \ W/m$

3 0

2 years ago

W²-5w+14

alexira [117]

Answer:w²-5w+14

4x²+11x+16

6x²+7x-10

x²-16x+56

Explanation:

7 0

2 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

3 years ago