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mixas84 [53]
3 years ago
11

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

complete. a. What is the radius of the curve that the plane follows in making the turn? b. What is the magnitude of the centripetal acceleration during the turn?
Physics
1 answer:
bulgar [2K]3 years ago
3 0

To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.

The perimeter of a circle is defined by

P = 2\pi r

Where,

r= radius

While centripetal acceleration is defined by

a=\frac{v^2}{r}

Where,

v= velocity

r= radius

PART A)

The distance of a body can be defined based on the speed and the time traveled, that is

x = v*t

For our values the distance is equal to

x = 15*115=1725m

The plane when going to make the turn from east to south makes a quarter of the circumference that is

\frac{P}{4} = \frac{2\pi r}{4}

The same route you take is the distance traveled, that is

x = \frac{P}{4}

x = \frac{2\pi r}{4}

1725 = \frac{2\pi r}{4}

r = 1098.17m

PART B)

With the radius is possible calculate he centripetal acceleration,

a = \frac{v^2}{r}

a = \frac{115^2}{1098.17}

a = 12.04m/s^2

Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of 12.04m/s^2

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Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17
Alexus [3.1K]

Answer:

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So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

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GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

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The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the
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Answer:

50 N

Explanation:

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F = (120 N / 0.4) / 6

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