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MissTica
2 years ago
14

Suppose you wish to whirl a pail full of water in a vertical circle at a constant speed without spilling any of its contents (ev

en at the top of the circle!). If your arm is 0.65 m long (from shoulder to fist) and the distance from the handle to the surface of the water is 19.0 cm, what minimum speed is required?
Physics
1 answer:
Yanka [14]2 years ago
6 0

Answer:

V = 2.87 m/s

Explanation:

The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.

Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:

Centripetal acceleration = V^2 / r

where r is the distance of water from the pivot or shoulder.

For our case, r will be 0.65 + 0.19 = 0.84 m

and solving the above equation we get:

9.81 = V^2 / 0.84

V^2 = 8.2404

V = 2.87 m/s

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The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

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2 years ago
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saul85 [17]

Answer:

<u>B</u>

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4 0
1 year ago
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A sample of glass has a mass of 15 g and takes up 14.3 cm 3 of space. What is the density? If you drop a piece of the same glass
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A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is
Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with <u>uniform circular motion</u>, her centripetal acceleration a_{c} is given by the following equation:  

a_{c}=\frac{V^{2}}{r} (1)

Where:  

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V is the<u> tangential speed</u>

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Isolating V from (1):

V=\sqrt{a_{c}r} (2)

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<u />

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