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liraira [26]
3 years ago
8

Your roommate is working on his bicycle and has the bike upside down. He spins the 60 cm

Physics
1 answer:
tigry1 [53]3 years ago
5 0
Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.

The pebble in the tread goes by 3 times every second.

This is the same as 3 times per second.

Recall the unit of frequency is Hertz or per second, s⁻¹

So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz

For angular motion:

Angular speed, ω = 2πf

                         = 2*π*3

                         = 6π   rad/s

Linear speed, v = ωr =  6π * 0.3 = 1.8π m/s


Linear acceleration, a = v² / r

                               a = 1.8π * 1.8π / 0.3 = 10.8π²   m/s²

Angular acceleration α = a/r  = 10.8π² / 0.3 = 36π² rad/s²


Angular speed = 6π rad/s ≈ 18.840 rad/s

The linear speed of the pebble = 1.8π  m/s ≈ 5.655 m/s

The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²

The linear acceleration of the pebble = 10.8π²  m/s ≈ 106.592 m/s²
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Answer:

True

Explanation:

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3 years ago
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
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Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

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