All categories

3 years ago

What are three automotive safety systems

Engineering

1 answer:

abruzzese [7]3 years ago

7 0

Answer:

Head-Up Display, Anti-Lock Braking Systems, and Electronic Stability Control.

Explanation:

You might be interested in

(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced

Nat2105 [25]

Answer:

Zx = 176In³

Explanation:

See attached image file

5 0

3 years ago

In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2

sp2606 [1]

Answer:

$l=24mm$

Explanation:

From the question we are told that:

Plane strain fracture toughness of $T=55 MPa-m1/2$

Y value $Y=1.0$

Stress level of $\sigma =200 MPa$

Generally the equation for length of a surface crack is mathematically given by

$l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2$

$l=\frac{1}{3.142}(\frac{55}{1*200})^2$

$l=0.024m$

Therefore

in mm

$l=24mm$

6 0

3 years ago

A team of engineers is working on a design to increase the power of a hydraulic lever. They have brainstormed several ideas. Whi

Goshia [24]

Answer:

Trye

Explanation:

7 0

3 years ago

A pump is used to deliver water from a lake to an elevated storage tank. The pipe network consists of 1,800 ft (equivalent lengt

Nataly_w [17]

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

7 0

3 years ago

Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from

kykrilka [37]

Answer:

Given:

high temperature reservoir $T_{H} =1000k$

low temperature reservoir $T_{L} =400k$

thermal efficiency $n_{1}= n_{2}$

The engines are said to operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

$n_{1} =1-\frac{T}{T_{H} }$

The thermal efficiency of second heat engine can be written as

$n_{2} =1-\frac{T_{L} }{T}$

The temperature of intermediate reservoir can be defined as

$1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k$

8 0

3 years ago