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2 years ago

Provide one example of a bad collision, and suggest an engineering solution to avoid the collision.

Engineering

1 answer:

JulsSmile [24]2 years ago

7 0

Answer:

1). Keep your distance. Drive far enough behind the car in front of you so you can stop safely. ...

Drive strategically. Avoid situations that could force you to suddenly use your brakes. ...

Don't get distracted. ...

Don't drive when drowsy or under the influence.

2). By far the deadliest accident type is the head-on collision. Head-on collisions consider both vehicle's speed at the time of the crash, which means even an accident at lower speeds can be catastrophic

Explanation:

first is how to avoid the collision and second is bad collision

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If Ori gives a friend three reasons for preferring soccer to basketball, that is an algorithm.

irinina [24]

Answer:

False I'm pretty sure sorry If its wrong

3 0

2 years ago

Read 2 more answers

A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu

bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0

3 years ago

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities <em> $n_{th}$ </em> therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

5 0

3 years ago

Find E[x] when x is sum of two fair dice?

Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

P(X=3)=P(1,2)+P(2,1)=

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

P(X=11)=P(5,6)+P(6,5)=

P(X=12)=P(6,6)=

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

⋅P(X

)

=2×

+3×

+4×

+5×

+6×

+7×

+8×

+9×

+10×

+11×

+12×

+1+

E(X

)=∑X

⋅P(X

)

=4×

+9×

+16×

+25×

+36×

+49×

+64×

+81×

+100×

+121×

+144×

+5+

+9+

121

987

329

=54.833

Then, Var(X)=E(X

)−[E(X)]

=54.833−(7)

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

5.833

=2.415

4 0

2 years ago

A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti

Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = $\frac{0.03194}{0.315}$

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m

here h is heat required and m is mass flow rate

H = 2680.36 × 0.10139

H = 271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = $\frac{H}{0.90}$

net boiler heat = $\frac{271.75}{0.90}$

net boiler heat = 301.94 kW

5 0

3 years ago