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spayn [35]
3 years ago
6

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick

is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation. With what period does it now oscillate?
Physics
1 answer:
rewona [7]3 years ago
3 0

Answer:

The new time period is  T_2 =  3.8 \  s

Explanation:

From the question we are told that

  The period of oscillation is  T =  5 \ s

   The  new  length is  l_2  =  0.76  \ m

Let assume the original length was l_1 = 1 m

Generally the time period is mathematically represented as

         T  =  2 \pi   \sqrt{ \frac{ I }{ mgh } }

Now  I is the moment of inertia of the stick which is mathematically represented as

           I  =  \frac{m * l^2 }{12 }

So

        T  =  2 \pi   \sqrt{ \frac{  m * l^2 }{12 *   mgh } }

Looking at the above equation we see that

        T  \ \ \  \alpha  \ \ \  l

=>    \frac{ T_2 }{T_1}  =  \frac{l_2}{l_1}

=>    \frac{ T_2}{5} =  \frac{0.76}{1}

=>     T_2 =  3.8 \  s

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Answer:

Explanation:

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R = N .

Ranking normal force from  highest  to smallest

150 N , 130 N , 120 N

B )

Frictional force is equal to the weight of the body because the body is held at rest .

Ranking of frictional force form largest to smallest

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Here frictional force is irrespective of the normal force acting on the body  because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .

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A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to
nadya68 [22]

Answer:

Explanation:

Given

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Time taken to fill the bucket t=50\ s

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\dot{m}=62.4\times 0.0267

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Answer: 1, 2, 6, 9

Explanation:

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ozzi

Answer: 74.1^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed=8.94\times 10^3 Joules

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c = specific heat capacity = 0.385J/g^0C

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Change in temperature ,\Delta T=T_f-T_i

Putting in the values, we get:

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T_f=74.1^0C

The final temperature of copper will be 74.1^0C

8 0
3 years ago
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