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dedylja [7]
3 years ago
5

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.
Physics
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:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

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By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0<em>.</em>

2\, q_{1} = q.

q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

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