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natta225 [31]
3 years ago
12

Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra

dius at which it had been rotated with the original spring, would the new period of rotation using the new spring be greater or less than the period of rotation using the original spring
Physics
1 answer:
malfutka [58]3 years ago
6 0

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the  period of rotation of the mass is mathematically represented as

       T = 2 \pi \sqrt{\frac{I}{k} }

Here I is the moment of inertia of the mass about the rotation axis and  k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

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Riley is cleaning his room(a once a year activity) and pulls the Hoover with a force of 8 N a total distance of 20 metres. How m
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160J

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A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
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Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

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A. Determination of the time.

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Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

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0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

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