All categories

4 years ago

What driving behavior can increase risk

Engineering

1 answer:

mario62 [17]4 years ago

6 0

Answer:

instances of hazardous practices are not utilizing headlights appropriately, driving unreasonably quick for conditions, closely following, dangerous passing, or path evolving, and so on. Inability to wear a safety belt is a high-chance conduct that frequently makes the results of an impact more terrible.

Explanation:

hope this helps :) !!

You might be interested in

The inner surface of a hollow cylinder is subjected to tangential and axial stresses of 40,000 and 24,000 psi, respectively. Det

Furkat [3]

Answer:

15,000 psi

Explanation:

The solution / solving is attach below.

5 0

3 years ago

Engineers, scientists, and designers innovate products by taking apart existing

masya89 [10]

Answer:

Mechanical dissection

Explanation:

Taking apart old things and innovating to improve upon them is mechanical dissection.

7 0

3 years ago

For goods-producing firms, at which of the following levels of resource planning does scheduling for individual subassemblies an

VashaNatasha [74]

Answer:

Disaggregation

Explanation:

In a company it is a way to create operational plans that are focused, either by time or by section.

3 0

3 years ago

Befor operating any plant,machinery or equipment,you should

SIZIF [17.4K]

Answer:

Always make sure that you are properly protected and that it is all clear to be operated on.

3 0

4 years ago

Read 2 more answers

A cantilever timber beam with a span of L = 4.25 m supports a linearly distributed load with maximum intensity of w0 = 5.5 kN/m.

9966 [12]

Answer:

the minimum width is b= 0.1414m = 141mm

Explanation:

]given,

L= 4.25

w₀ = 5.5kN/m,

allowable bending stress = 7MPa

allowable shear stress = 875kPa

h/b = 0.67

b = ?

for a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m,

the maximum moment, M exerted by the timber is = $\frac{w₀ L²}{9√3}[/texM = w₀ L²}/{9√3 = 99.34/15.6 =6.367kNmfor a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m, the shear force, V = [tex]\frac{w₀ L}{2}[/texV = {w₀ L}/{3} = 7.79kNfor maximum bending stress of a rectangular timber, B, = [tex]\frac{6M}{bh²}$

given h/b = 0.67, i.e h=0.67b

allowable bending stress = $\frac{6M}{bh²}$ = 7000kPa

7000 = (6*6.37)/ (b *(0.67b)² ) = 38.21/0.449b³

3080b³=38.21

b³ = 38.21/3080 = 0.0124

b = 0.232m

h=0.67b = 0.67* 0.232 = 0.155m

for allowable shear stress = (3V)/(2bh)

875 = (3V)/(2bh) = (3x 7.79)/(2xbx0.67b)

875 = 23.375/1.34b²

1172.5 b²= 23.375

b² =0.0199

b= 0.1414m

h=0.67b = 0.67* 0.1414 = 0.095m

the minimum width is b= 0.1414m = 141mm

4 0

3 years ago