3 years ago

You could be sued if you injure someone while rescuing them if...

2 answers:

4 0

It’s 4. One is accidental and can be the fault of the car dealership or maker, same with two. Three is their fault, and you could sue them.

navik [9.2K]3 years ago

3 0

I believe number 4 I could be wrong but I think it’s 4

You might be interested in

PLEASE ANSWER FOR DRIVERS ED! WILL GIVE BRAINLIEST

Zepler [3.9K]

Answer:D

Explanation:

Google it it’s 100 ft

8 0

3 years ago

we want to make a schottky diode on one surface of an n-type semiconductor, and an ohmic contact on the other side. the electron

RoseWind [281]

Answer:

<h2>hope it helps you see the attachment for further information .....✌✌✌✌✌</h2>

7 0

4 years ago

Controlling your vehicle

Ratling [72]

Answer:

5. D

6. c

7. d

Explanation:

3 0

3 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

3 years ago

A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest

sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

$y = ax^{2} +bx+c\\dy/dx=2ax+b\\$

At highest point of the curve $dy/dx=o$

$2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275$

$-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\$

Station PVT

$Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)$

3 0

3 years ago