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3 years ago

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Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical dist

ance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m.

1 answer:

WINSTONCH [101]3 years ago

6 0

Answer:

a)

2474 N

b)

148438 N

Explanation:

(a)

v₀ = initial velocity of the car = 90 km/h = 25 m/s

v = final velocity of the car = 0 m/s

d = stopping distance for the car = 120 m

F = force needed to bring the car to rest

m = mass of the car = 950 kg

Using Work-change in kinetic energy theorem

F d Cos180 = (0.5) m (v² - v₀²)

F (120) (-1) = (0.5) (950) (0² - 25²)

F = 2474 N

b)

v₀ = initial velocity of the car = 90 km/h = 25 m/s

v = final velocity of the car = 0 m/s

d = stopping distance for the car = 2 m

F = force needed to bring the car to rest

m = mass of the car = 950 kg

Using Work-change in kinetic energy theorem

F d Cos180 = (0.5) m (v² - v₀²)

F (2) (-1) = (0.5) (950) (0² - 25²)

F = 148438 N

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Answer:

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Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

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Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

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Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

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Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

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$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

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3 years ago

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