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Readme [11.4K]
4 years ago
10

The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2

m long is used to rotate a 0.5kg object in a circle on a frictionless table with the string horizontal. The maximum speed that the mass can attain under these conditions without the string breaking is most nearly:
Answer is 20 m/s
Please show me how I can get the answer. Thank you.
Physics
1 answer:
vichka [17]4 years ago
0 0

Answer:

Explanation:

Think about circular motion when answering this question. First determine the maximum force that can be applied on the string. F = mg so F = (10)(10) = 100 N. Then determine the centripetal acceleration of the .5 kg mass, a = F/m so a = 100/.5 = 200 m/s².  On the equation sheet, use equation a(centripetal acceleration) = v²/r so 200 = v²/2 therefore v = 20 m/s. Hope this helps!

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A ball is thrown upwards from the edge of a cliff. The horizontal velocity and vertical velocity are both 20m/s the distance fro
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Answer:

20m

6.9s

Explanation:

The vertical velocity of the ball is 20m/s. We can calculate the kinetic energy which gets transferred to potential energy once it gets to the top.

E_k = E_p

0.5mv^2 = mgh

h = \frac{0.5v^2}{g}

we can subtitute v = 20m/s and g = 10m/s2

h = \frac{0.5*20^2}{10} = 20 m

So the ball could go 20m high from the child hand, or 120m fro the bottom of the cliff.

The time it takes for the ball to travels to the top is the time it takes for it to decelerate from 20m/s to 0m/s with gravitational deceleration g = 10m/s2

t = v / g = 20 / 10 = 2s

Then the ball will start accelerating down ward with a constant acceleration of g = 10m/s. In order to cover distance d of 120m from the top to the bottom of the cliff

d = \frac{gt_2^2}{2}

t^2 = \frac{2d}{g} = \frac{2*120}{10} = 24

t = \sqrt{24} = 4.9s

So the total time it takes is 4.9 + 2 = 6.9s

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3 years ago
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Use this technique to find a formula for the intensity I of a sound, in terms of the sound level β and the reference intensity I
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Answer:

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