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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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Ride in a car down the street: constant velocity,
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uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work i
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Answer:

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(b) The  force will stretch the spring by 3.8 cm

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W = ¹/₂kx²

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work done is given by;

W = ¹/₂kx²

W = ¹/₂ (266.67)(0.02)²

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x = 0.0375 m

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A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a
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d= 7.32 mm

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πr² = 42.14 mm²

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d= 7.32 mm

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