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MAXImum [283]
3 years ago
12

In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c

an alter the net force on a passenger. Consider a passenger of mass m riding around a horizontal circle of radius rat speed v. What is the variation dF in the net force magnitude for(a) a variation dr in the radius with ν held constant,(b) a variation dv in the speed with r held constant, and(c) a variation dT in the period with r held constant?
Physics
1 answer:
Bess [88]3 years ago
8 0

Answer:

Part a)

dF = -\frac{mv^2}{r^2} dr

Part b)

dF = \frac{2mvdv}{r}

Part c)

dT = - \frac{2\pi r}{v^2} dv

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

F = \frac{mv^2}{r}

now variation in force is given as

dF = -\frac{mv^2}{r^2} dr

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

F = \frac{mv^2}{r}

so we have

dF = \frac{2mvdv}{r}

Part c)

As we know that time period of the circular motion is given as

T = \frac{2\pi r}{v}

so here if radius is constant then variation in time period is given as

dT = - \frac{2\pi r}{v^2} dv

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An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
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To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

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A 4.5cm tall object is placed 28cm in front of a spherical mirror. It is to produce a virtual image that is upright and 3.5cm ta
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3 years ago
the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter
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Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

        PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

I                           0                0.10

C                         +S               +2S

E                          S              0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 × 10⁻⁸ = S . (0.10)²

S = 3.6 × 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L

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3 years ago
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