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tiny-mole [99]
3 years ago
8

Alcohol consumption slows people's reaction times.In a controlled government test,it takes a certain driver 0.320 s to hit the b

rakes in a crisis when unimpaired and 1.00 s when drunk.When the car is initially traveling at 90km/h,how far does the car travel before coming to a stop when the person is drunk compared to sober?
Physics
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

When the person is drunk the car travels 20.1 meters more than when the person is sober. The distance traveled by car is:

Person drunk: 28.1 m

Person sober: 8 m     

Explanation:

The distance traveled can be found using the following equation:

v = \frac{x}{t} \rightarrow x = v*t

Where:

v: is the speed = 90 km/h

t: is the time

When the person is sober, the time that takes to hit the brakes is 0.320 s, so we have:

x = v*t = 90 \frac{km}{h}*\frac{1 h}{3200 s}*0.320 s = 8 \cdot 10^{-3} km = 8 m

And when the person is drunk, the time is 1.00 s, hence the distance is:

x = v*t = 90 \frac{km}{h}*\frac{1 h}{3200 s}*1 s = 0.0281 km = 28.1 m

The distance traveled by the car when the person is drunk compared when the person is sober is:

d = 28.1 m - 8 m = 20.1 m

Therefore, when the person is drunk the car travels 20.1 meters more than when the person is sober.  

I hope it helps you!

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ikadub [295]

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

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3 years ago
What is one joule work
natta225 [31]
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8 0
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5 0
2 years ago
An average man is 80 k (800 N) and the area of the shoes he is wearing is 0.0092 m2. What is the pressure he is exerting on the
S_A_V [24]

Answer:

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Explanation:

Data:

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  • A = 0.0092 m²
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Use the formula:

  • \boxed{\bold{P=\frac{F}{A}}}

Replace and solve:

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The pressure it exerts on the ground is <u>86956.52 Pascal.</u>

Greetings.

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