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3 years ago

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An automobile moves at a constant speed over the crest of a hill traveling at a speed of 88.5 km/h. At the top of the hill, a pa

ckage on a seat in the rear of the car barely remains in contact with the seat. What is the radius of curvature (m) of the hill?

1 answer:

Sergio039 [100]3 years ago

3 0

Answer:

$r=61.65m$

Explanation:

Since the package remains in contact with the car's seat, the package's speed is equal to the car's speed. At the top on the mountain the package's centripetal force must be equal to its weight:

$mg=F_c$

The centripetal force is defined as:

$F_c=ma_c=\frac{mv^2}{r}$

Here v is the linear speed of the object and r is the radius of curvature. We need to convert the linear speed to $\frac{m}{s}$ :

$88.5\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=24.58\frac{m}{s}$

Now, we calculate r:

$mg=\frac{mv^2}{r}\\r=\frac{v^2}{g}\\r=\frac{(24.58\frac{m}{s})^2}{9.8\frac{m}{s^2}}\\\\r=61.65m$

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part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

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part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

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mrs_skeptik [129]

Answer:

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