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2 years ago

10

A manufacturer has been asked to produce 100 customized metal discs with a particular pattern engraved on them. Which production

process should be selected?

2 answers:

topjm [15]2 years ago

8 0

I’m just here for points because I have test and I need them lol

saul85 [17]2 years ago

3 0

Answer:

Job Shop production

Explanation:

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Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio

SVETLANKA909090 [29]

This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft

Explanation:

Given that;

section W 21 x 73 steel beam;

now from the steel table table for this section;

Zx = Sx = 151 in³

also given that; fy = 50 ksi and Cb = 1.0

QMn = 0.9 × Fy × Zx

so we substitute

QMn = 0.9 × 50 × 151

QMn = 6795 k-inch

we know that;

12inch equals 1 foot

so

QMn = 6795 k-inch / 12

QMn = 566.25 f-ft

Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft

7 0

3 years ago

A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

Kazeer [188]

To solve this problem we will apply the concepts related to translational torque, angular torque and the kinematic equations of angular movement with which we will find the angular displacement of the system.

Translational torque can be defined as,

$\tau = Fd$

Here,

F = Force

d = Distance which the force is applied

$\tau = (1N)(1m)$

$\tau = 1N\cdot m$

At the same time the angular torque is defined as the product between the moment of inertia and the angular acceleration, so using the previous value of the found torque, and with the moment of inertia given by the statement, we would have that the angular acceleration is

$\tau = I\alpha$

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{1N\cdot m}{100kg\cdot m^2}$

$\alpha = 0.01rad/s^2$

Now the angular displacement is

$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$

Here

$\omega_0$ = Initial angular velocity

t = time

$\alpha =$ Angular acceleration

$\theta$ = Angular displacement

Time is given as 1 minute, in seconds will be

$t = 1m = 60s$

There is not initial angular velocity, then

$\theta= \frac{1}{2}\alpha t^2$

Replacing,

$\theta= \frac{1}{2}(0.01)(60)^2$

$\theta = 18rad$

The question neglects the effect of gravitational force.

4 0

3 years ago

A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t

maksim [4K]

Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

A) The flow velocity intensity for the input radial coordinate "r" is given by;

U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

Thus,

U(r) = (V_c)[1 - (2r/D)²]

From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m

But we are to find the velocity at the centre of the tube, thus;

We will use the radius across the horizontal distance which will be;

0.05 - 0.012 = 0.038m

Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

Thus;

U(r) = (V_c)[1 - (2r/D)²]

0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]

0.8 = (V_c)[1 - (0.76)²]

V_c = 0.8/0.4224 = 1.894 m/s

B) flow rate is given by;

ΔV = Average Velocity x Area

Now, average velocity = V_c/2

Thus, average velocity = 1.894/2 = 0.947 m/s

Area(A) = πr² = π x 0.05² = 0.007854 m²

So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s

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3 years ago

For three control stations,there should be how many start buttons in parallel with the suxiliary contact

valentinak56 [21]

The three load contacts connected between the three-phase power line and the motor close to connect the motor to the line. The normally open auxiliary contact connected in parallel with the two Start buttons closes to maintain the circuit to M coil when the Start button is released.

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3 years ago

What is -4 (negative 4) in a 2's complement of 8 bits?

Pachacha [2.7K]

Answer: Yes

Explanation: -4 x 2 = 8

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3 years ago