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3 years ago

Explain what a margin of safety is in driving as well as how it can help minimize risk.

Engineering

1 answer:

Yakvenalex [24]3 years ago

6 0

Answer:

A safety margin is the space left between your vehicle and the next to provide room, time and visibility at every instant

Explanation:

A safety margin is defined as an allowance given between your vehicle and the next vehicle in front to provide enough room, visibility and time to move in a safe manner to prevent the occurrence of an accident at anytime the frontal vehicle suddenly stops or slows down

Safety margins help minimize risks in the following way

1) A common knowledge of safety margins, improves predictability among road users, thereby minimizing the risk traffic accidents caused due to late communication

2) The use of safety margins helps minimize the risk due to a change in driving conditions such as when the road becomes more slippery from being covered with fluid that is being wetted

3) Safety margin can help prevent the occurrence of an accident between vehicles due to failure of a car system, such as a punctured tire or failed breaking system

4) Safety margin helps to protect road users from the introduction of obstacles on the main roads such as ongoing road construction, broken down vehicles, road blockage by vehicles involved in an accident etc

5) Safety margin help protect road users from being involved in an accident due to the loss of driving focus of the driver of the frontal vehicle

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A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled

Vikentia [17]

Answer:

Force per unit plate area is 0.1344 $N/m^{2}$

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, $\mu =1.92\times 10^{- 3} Pa-s$

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

$F = \tau A$

where

$\tau$ = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = $(\tau ' +\tau '')A$

$\frac{F}{A} = \tau ' +\tau ''$

Also

F = $\frac{\mu v}{d}$

where

$\mu$ = dynamic coefficient of viscosity

Pressure, P = $\frac{F}{A}$

Therefore,

$\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}$

$\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}$

8 0

3 years ago

Me ayudas plis noce

alekssr [168]

Answer:

Explanation:

7 0

2 years ago

Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using

VashaNatasha [74]

Answer:

For any string, we use $s = xyz$

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

Consider any string a i b j c k in the language. If i = 1 or i > 2, we take $x = \epsilon$ and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.

For i = 2, we can take and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
Finally, for the case i = 0, we take $x = \epsilon$ , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.

8 0

3 years ago

If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det

Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

solution:

to find the shear stress

T_allow=T_fail/F.S

=120 Mpa/2.5

=48 Mpa

we know that,

V=P

Area for shear head:

A(head)=π×d×t

=π×0.04×0.075

=0.003×πm^2

Area for plate:

A(plate)=π×d×t

=π×0.08×0.03

=0.0024×πm^2

now we have to find shear stress for both head and plate

For head:

T_allow=V/A(head)

48 Mpa=P/0.003×π ..(V=P)

P =48 Mpa×0.003×π

=452.16 KN

For plate:

T_allow=V/A(plate)

48 Mpa=P/0.0024×π ..(V=P)

P =48 Mpa×0.0024×π

=361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

P=361.91 KN

note:

find the attached pic

7 0

3 years ago

3. In order to obtain your commercial driver's license (CDL) you must first:

Murljashka [212]

A and C is the answer to the question. Be 15 years old & get a permit

8 0

3 years ago