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denpristay [2]
3 years ago
10

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 12 blades and rotates at a

n angular speed of 1.35 rad/s. The opening between successive blades is equal to the width of a blade. A golf ball (diameter 4.50 10-2 m) has just reached the edge of one of the rotating blades (see the drawing). Ignoring the thickness of the blades, find the minimum linear speed with which the ball moves along the ground, such that the ball will not be hit by the next blade
Physics
1 answer:
tester [92]3 years ago
7 0

Answer:

v<em>min</em> = 0.23 m/s

Explanation:

The golf ball must travel a distance equal to its diameter in the time between blade arrivals to avoid being hit. If there are 12 blades and 12 blade openings and they have the same width, then each blade or opening is 1/24 of a circle of is 2π/24 = 0.26 radians across.

Therefore, the time between the edge of one blade moving out of the way and the next blade moving in the way is

time = angular distance/angular velocity

⇒ t = 0.26 rad / 1.35 rad/s = 0.194 s

The golf ball must get completely through the blade path in this time, so must move a distance  equal to its diameter in 0.194 s, therefore the speed of the golf ball is

v =d/t

⇒ v = 0.045 m / 0.194 s = 0.23 m/s

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How do we get this equation ??<br><br><br> H=V^2÷R
lesantik [10]

H=\frac{V^2}{R}\times t is the equation that represents the Joule's law of heating.

<h3>Explanation:</h3>

Joule's law of heating defines the heat generated by any current flowing conductor is directly proportional to  

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2. Resistance of the conductor (R)

3. Time for which current is passed (t)

Hence, Heat generated = H = I^2 Rt .....................(1)

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V\ \alpha\ I\\V=I\times R ...............................................(2)

From (2), Current (I) can be rewritten as

I = \frac{V}{R} ........................................................(3)

Substituting (3) in (1), we get

H = I^2\times R\times t \\=(\frac{V}{R} )^2\times R\times t\\\\=\frac{V^2}{R^2}\times R\times t = \frac{V^2}{R}\times t\\\\ H =\frac{V^2}{R}\times t

3 0
3 years ago
A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving a
skelet666 [1.2K]

Answer:0.077

Explanation:

Given

banked designed for traffic moving at 58 km/h\approx 16.11 m/s

Radius of the curve 201 m

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For banking of road

tan\theta =\frac{v^2}{rg}

tan\theta =\frac{16.11^2}{201\times 9.81}

\theta =7.49^{\circ}

Centripetal acceleration is given by

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

Taking component of centripetal acceleration

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a_{perpendicular}=\frac{v^2sin\theta }{r}

From FBD

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where f_s is frictional force

N-mgcos\theta =ma_{perpedicular}

N=mgcos\theta +ma_{perpedicular}----2

and we know coefficient of friction is given by

\mu =\frac{f_s}{N}

\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}

\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}

\mu =\frac{1.2804-0.5202}{9.726+0.068}

\mu =0.077

7 0
3 years ago
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