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MissTica
3 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]3 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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3 years ago
An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q
Marina CMI [18]

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency f  = ( \dfrac{1}{2 \pi}) \omega

where;

\omega = \sqrt{\dfrac{k}{m} }

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{k}{m} }  \Bigg )

However;

k  = \dfrac{F}{x} and;

mass m = m_{car } + m_{person}

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} }  \Bigg )

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} }  \Bigg )

where;

F = m_{person}g

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} }  \Bigg )

replacing the values;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} }  \Bigg )

\mathbf{f = 0.442 \ Hz}

8 0
3 years ago
A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s
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Answer:

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(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final - K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

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The collective body of observations of a natural phenomenon on which scientific explanations are based is called?
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Experimental Evidence.

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