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leva [86]
3 years ago
12

Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their

center. A rider sits at the outer edge of the disk and holds onto a metal bar while someone pushes on the ride to make it rotate. Suppose that a typical time for one rotation is 6.0 s and the diameter of the ride is 12 ft .
Physics
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

Though the question is not specified here, but this information can determine the following quantity: period T= 6 secs, Frequency F=1/6 Hz, speed of rotation V= 2 pi ft/sec and wave length =pi/3 ft

Explanation:

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find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds
Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}

Final speed of the train, v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}

Displacement of the train, S=234\textrm{ km}=234\times 1000=234000\textrm{ m}

Using Newton's equation of motion,

v - u = at\\a=\frac{v-u}{t}

Now, using Newton's equation of motion for displacement,

v^{2}-u^{2}=2aS

Now, plug in the value of a=\frac{v-u}{t} in the above equation. This gives,

v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}

Now, plug in 234000 m for S, 25 m/s for v and 20 m/s for u. Solve for t.

t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}

Therefore, the time taken by the train is 10400 s.

3 0
4 years ago
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