If you see rough patches, loose gravel, or potholes on the road, you should _____

Anytime scaffolds are assembled or __________, a competent person must oversee the operation.

egoroff_w [7]

D. made because made is a synonym for assembled

4 0

2 years ago

Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in

salantis [7]

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

class Program

{

static void Main(string[] args)

{

int[] Number = new int[10]; // Declare array of 10 elements

//Accept Input

int counter = 0;

while(counter<10)

{

Console.WriteLine("Enter Number " + (counter + 1)+": ");

string var = Console.ReadLine();

Number[counter] = Convert.ToInt32(var);

counter++;

}

//Initialize largest to first element of the array

int Largest = Number[0];

//Determine Largest

for(int i=0;i<10;i++)

{

if(Largest < Number[i])

{

Largest = Number[i];

}

//Print Largest

Console.WriteLine("The largest input is "+ Largest);

Console.ReadLine();

}

8 0

2 years ago

A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t

Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

$\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa$

The design factor for metals is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}$

The design factor for plastic and ceramics is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}$

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17$

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09$

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125$

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12$

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039$

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26$

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

$DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5$

This material will break.

3 0

2 years ago

The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr

Rashid [163]

Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below; if we take an integral look at the component of force in the direction of the displacement; we have

$F_x=$ Fcos 40°

$F_x=$ 100 (cos 40°)

$F_x=$ 76.60 N

Workdone by the friction force can now be determined as:

W = $F_x$ × displacement

W = 76.60 × 2

W = 153.2 J

∴ the work done by the friction force = 153.2 J

7 0

2 years ago

An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t

11Alexandr11 [23.1K]

Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

$E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J = 3.29* 10^{-8} eV$ .

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

λ = c/v

dλ/dv = -c/v² = -λ²/c

Therefore,

∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

= 9.20 × 10⁻¹⁵ m or 9.20 fm

The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m

6 0

2 years ago

If you see rough patches, loose gravel, or potholes on the road, you should ______.l