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2 years ago

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Time left 0:35:32 Three steel rod (E = 200 GPa) supports 36 KN Load P. Each of the rods AB and CD has a 200 mm? cross- sectional

area and rod EF has 625 mm2 cross-sectional area. C D Compute for the stress in the rod. AB. a. 15.25 MPa b. 38.10 MPa C 19.05 MPa d. 30.50 MPa

1 answer:

taurus [48]2 years ago

8 0

Answer:

tama po ba

Explanation:

correct me if im wrong

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Concerned with the number of maintenance visits the rocket can undergo before being out of service, you have been informed that

Ainat [17]

Answer:

(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

Explanation:

Solution

(a) Let Mn denotes the number of maintenance visits after the nth journey

Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3

We therefore, notice that M follows an arithmetic sequence

So,

Mn = M₁ + (n-1) (M₂ -M₁)

= 1 + (n- 1) 1 = n

or Mn =n

(b) For what value of n we will get fro Mn > 10

Thus,

n > 10 (exceed 10) or n =11

(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused

So,

n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

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The purpose of daytime running lights (DRLs) is to _____.

balu736 [363]

D make your car easy to spot in daytime because they do not illuminate a lot

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4 years ago

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Write multiple if statements:

lora16 [44]

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3 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

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KengaRu [80]

Answer:

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Explanation:

A skilled worker with the ability to operate computer numerically controlled (CNC) machines is qualified to work in a machinist position.

A machinist is a person who is properly skilled and consists of advanced knowledge regarding the functions of a CNC machine. He can use different mechanisms and complex numerical functions of the machine to carry out different tasks. Any person who lacks the official learning of mechanisms cannot operate such machines effectively.

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