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Katyanochek1 [597]
3 years ago
8

A book with a mass of 3.3 kg sits on a bookshelf. If it has a gravitational

Physics
1 answer:
erma4kov [3.2K]3 years ago
4 0
<h2><u>AnswEr :</u> </h2>

Given that,

  • Mass of the Book (M) = 3.3 Kg

  • Gravitational Potential Energy (P) = 57 J

  • Acceleration Due to Gravity (g) = 9.8 m/s²

We have to find the height of the shelf

We know that,

\tt P = MgH

Putting the values,

\sf \: 57 = 3.3 \times 9.8 \times h \\  \\  \implies \:  \sf \: 57 = 32.34 \times h \\  \\  \implies \:  \sf \: h = 1.76 \: m \: (approx)

<h3>The height of the shelf is 1.76 m </h3>
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A thin soap bubble of index of refraction 1.33 is viewed with light of wavelength 550.0nm and appears very bright. Predict a pos
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Answer:

The possible thickness of the soap bubble = 1.034\times 10^{-7}\ m.

Explanation:

<u>Given:</u>

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2\mu t=\left (0+\dfrac 12\right )\lambda\\t=\dfrac{\lambda}{4\mu}\\\ \ = \dfrac{550\times 10^{-9}}{4\times 1.33}\\\ \ = 1.034\times 10^{-7}\ m.

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To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
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To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

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a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

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v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

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Calculate the time needed by the motorcycle to reach an acceleration of

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a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

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PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

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Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

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PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

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