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Lorico [155]
3 years ago
13

The shape of the orbit for most comets is a(n): circle parabola ellipse oval

Physics
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

Ellipse

Explanation:

Most comets have an elliptical path of orbit.

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The finished product or waste that forms as a result of a process is known as _______.
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The finished product or waste that forms as a result of a process is known as by-product.
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A wave travels at a constant speed.How does the frequency change if the wavelength is reduced by a factor of 3 The frequency dec
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Answer:

The frequency increases by a factor of 3.

Explanation:

The relation between speed, wavelength and frequency of a wave is given by :

v=f\lambda

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f\propto \dfrac{1}{\lambda}

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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}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

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

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

27.5625 = 1.21 + \frac{v^4}{122500}

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

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

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

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net 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}

v=48.2796ft/s

PART C)

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

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

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}

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3 years ago
What is the kinetic energy of a 0.25kg ball rolling at a speed of 2.5m/s
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Answer:

Explanation:

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KE = 1/2(0.25kg)(2.5m/s)²

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