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ser-zykov [4K]
3 years ago
15

Which of the following factors does not promote safety in the shop?

Engineering
1 answer:
sergey [27]3 years ago
6 0
“Thinking about pleasant things to pass the time” would not promote safety in the shop because it would be taking the focus away from important tasks, which in turn decreases safety.
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The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

3 0
3 years ago
A construction manager is looking for workers to build a series of walkways within a large garden apartment. What type of workme
Lesechka [4]

Answer:

<u>construction workers</u>

Explanation:

A construction worker is usually someone with the technical skills and abilities needed to manually construct physical infrastructures.

Since a walkway is an infrastructure, hiring skilled construction workers should perform the project of building the series of walkways within the large garden apartment.

3 0
3 years ago
3. Suppose you work for this company that evaluates Boolean circuits in exponential time. Since you are very smart, your manager
hammer [34]

Answer:

Ergr5

Explanation:

3 0
3 years ago
Water is being added to a storage tank at the rate of 500 gal/min. Water also flows out of the bottom through a 2.0-in-inside di
melomori [17]

Answer:

From the answer, the water level is falling (since rate of outflow is more than that of inflow), and the rate at which the water level in the storage tank is falling is

(dh/dt) = - 0.000753

Units of m/s

Explanation:

Let the volume of the system at any time be V.

V = Ah

where A = Cross sectional Area of the storage tank, h = height of water level in the tank

Let the rate of flow of water into the tank be Fᵢ.

Take note that Fᵢ is given in the question as 500 gal/min = 0.0315 m³/s

Let the rate of flow of water out of the storage tank be simply F.

F is given in the form of (cross sectional area of outflow × velocity)

Cross sectional Area of outflow = πr²

r = 2 inches/2 = 1 inch = 0.0254 m

Cross sectional Area of outflow = πr² = π(0.0254)² = 0.00203 m²

velocity of outflow = 60 ft/s = 18.288 m/s

Rate of flow of water from the storage tank = 0.0203 × 18.288 = 0.0371 m³/s

We take an overall volumetric balance for the system

The rate of change of the system's volume = (Rate of flow of water into the storage tank) - (Rate of flow of water out of the storage tank)

(dV/dt) = Fᵢ - F

V = Ah (since A is constant)

dV/dt = (d/dt) (Ah) = A (dh/dt)

dV/dt = A (dh/dt) = Fᵢ - F

Divide through by A

dh/dt = (Fᵢ - F)/A

Fi = 0.0315 m³/s

F = 0.0371 m³/s

A = Cross sectional Area of the storage tank = πD²/4

D = 10 ft = 3.048 m

A = π(3.048)²/4 = 7.30 m²

(dh/dt) = (0.0315 - 0.0370)/7.3 = - 0.000753

(dh/dt) = - 0.000753

4 0
3 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
jok3333 [9.3K]

Answer:

Q=4.98\times 10^{-3}\ m^3/s.

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

\mu =0.00164\ Pa.s

We know that

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where Q is volume flow rate

L is length of tube

d_i is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s.

6 0
3 years ago
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