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3 years ago

What braking system would be the most unsafe for a car going over 200mph?

Engineering

1 answer:

nasty-shy [4]3 years ago

8 0

Answer:

Deactivated Anti-Lock Braking System

Explanation:

Well this actually depends on the vehicles braking function

I'd say in this case not sure if ABS coz it prevents the vehicle from slipping while braking...

On the otherside ESP is inportant for it stability

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Các đặc điểm chính của đường dây dài siêu cao áp .

rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0

3 years ago

Which of the following is NOT part of a car's drive train?

FromTheMoon [43]

Answer:

B. rotor

Explanation:

The correct answer Is rotor because the others are part of a cars drivetrain

3 0

3 years ago

Read 2 more answers

Seawater containing 3.50 wt% salt passes through a series of 11 evaporators. Roughly equal quantities of water are vaporized in

statuscvo [17]

Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

Explanation:

F, W and B are the fresh feed, brine and total water obtained

w = 2 x 10^4 L/h

we know that

F = W + B

we substitute

F = 2 x 10^4 + B

F = 20000 + B .................EQUA 1

solute

0.035F = 0.05B

B = 0.035F/0.05

B = 0.7F

now we substitute value of B in equation 1

F = 20000 + 0.7F

0.3F = 20000

F = 20000/0.3

F = 66666.67 kg/hr

B = 0.7F

B = 0.7 * F

B = 0.7 * 66666.67

B = 46,666.669 kg/hr

the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

8 0

4 years ago

What did the US and USSR agree on in the INF Treaty? They agreed to reduce nuclear weapons. They agreed that new European nation

faust18 [17]

Answer:

they agreed to reduce nuclear weapons

4 0

3 years ago

Read 2 more answers

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

$F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}$

Substituting figures in the equation we have:

$= \frac{1}{1+0.1(2.5-1)+0(2-1)}$

= 0.75

Let's now calculate equivalent flow rate of the car using:

$Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}$

$= \frac{7500}{0.9*4*0.75*1}$

= 2777.7 pc/h/en

Calculating for traffic density, we have:

$D = \frac{Vp}{FFS}$

$D = \frac{2777.7}{70}$

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0

3 years ago