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3 years ago

8

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was

μs = 1. around 1962, three companies independently developed racing tires with coefficients of 1.6. this problem shows that tires have improved further since then. the shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. ]

1 answer:

True [87]3 years ago

5 0

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:

Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².

For a constant acceleration, distance = 402.3

m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2

µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

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Read 2 more answers

3. Using the F, m, a triangle, calculate the boy's mass. Use the "force

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3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

<u></u>

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

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3 years ago