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vladimir1956 [14]
2 years ago
12

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.7m. If the thrower takes 1.2s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Physics
1 answer:
harina [27]2 years ago
6 0

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

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In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to
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Answer:

17304 J

Explanation:

Complete statement of the question is :

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .

Part A

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Solution :

T = tension force in the tow rope = 350 N

L  = length of the incline surface = 120 m

W_{t}  = work done by tension force = ?

The tension force acts parallel to incline surface, hence work done by tension force is given as

W_{t} = T L\\W_{t} = (350) (120)\\W_{t} = 42000 J

h = height gained by the rider = 30 m

m = total mass of rider and tube = 84 kg

Potential energy gained is given as

U = mgh\\U = (84) (9.8) (30)\\U = 24696 J

Q  = Thermal energy created

Using conservation of energy

Q = W_{t} - U\\Q = 42000 - 24696\\Q = 17304 J

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