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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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Answer:

d. 0.3 N

Explanation:

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The S.I unit of Force is Newton (N).

Mathematically Force can be represented as,

F = ma .................. Equation 1

Where F = force, m = mass, a = acceleration.

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Where v = final velocity, u = initial velocity, t = time.

Given: v = 0.60 m/s, u = 0 m/s ( From rest), t= 0.16 s.

Substitute into equation 2

a = (0.60-0)/0.16

a = 3.75 m/s²

Also given: m = 80 g = 0.08 kg.

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F = 0.08(3.75)

F = 0.3 N.

Hence he average force = 0.3 N

The right option is d. 0.3 N

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3 years ago
the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located
vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector <em>E= </em>5.5 ×10¹³N/C(-x direction)

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3 years ago
A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ
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Answer:

The correct question is:

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The energy gained by a charged particle accelerated through a potential difference is given by

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And since \Delta V=100 V

The energy gained by the proton is

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