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Naya [18.7K]
3 years ago
8

A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th

e average speed (in m/s) of the blade tip in the helicopter's frame of reference. m/s
(b) What is its average velocity (in m/s) over one revolution? m/s
Physics
1 answer:
NeX [460]3 years ago
8 0

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

v=r\omega

v = 4.50\times110\times\dfrac{2\pi}{60}

v=51.83\ m/s

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

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mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

4 0
2 years ago
A constant force of 5.00 N acts on a 2.50 kg object for 10.0 s. What are the changes in the object’s momentum and velocity?
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Hope this answer helps, cause Idk, I might be wrong, but I still, I used the correct formulas, so I might be correct

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If the temperature warning light/gauge goes on, what should one do?
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A golfer badly misjudges a putt, sending the ball only one-quarter of the distance to the hole. The original putt gave the ball
IRISSAK [1]

Answer:

v = 2 v₀

Explanation:

For this exercise we will use the relationship between work and the change in kinetic energy

          W = ΔK

Where the work is of the friction force, which is always opposed to the movement whereby the angle is 180º

          W = - fr x

The distance for the first case is

         x = 1/4 L

         

We substitute

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        fr L = 2 m v₀²

In the second case, the new speed takes the ball the entire distance

        x = d

        -fr L = ½ m v²

     

We equal the two equations

        2 m v₀² = ½ m v²

        v² = 4 v₀²

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3 0
3 years ago
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