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katen-ka-za [31]
3 years ago
12

Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t

o launch an object horizontally with a speed of 2.08 m/s. If the object were attached to the spring, at what angular frequency (in rad/s) would it oscillate?
Physics
1 answer:
padilas [110]3 years ago
4 0

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

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\huge \mathfrak \red{Answer} \\  \\  \:   \:  \implies \: \sf 35 \: watt \\  \\  \\  \bf \: Given \:  :  \\  \\  \:  \:  \tt Q_{1} = 263j \\    \:   \tt \:  T_{1} =  - 10    {}^{ \circ} C \:  =  - 10 + 273 = 263 \: k \\  \:  \tt \:  T_{2}  = 25^{ \circ} C = 25 + 273 = 298 \:k \\  \\  \\  \bf \: To \: Find \:   :   \: \:  P ?\:  \\  \\ \:  \sf \green{Solution :} \\  \\  \:  \:   \large \: \bf \: P = Q _{2} -  Q_{1} \\  \\   \:  \:  \: \tt \: \rightarrow \frac{Q_{2}}{Q_{1}}  =  \frac{T_{2}}{T_{1}}  \\  \\  \:\rightarrow  \tt \:  \: \:   Q_{2} =  \frac{T_{2}}{T_{1}}  \times Q_{1} \\  \\  \:  \tt \:  \: \rightarrow \:  Q_{2}=  \frac{298}{ \cancel 263}  \times \cancel 263  \\   \\ \:   \: \tt \:\rightarrow \: Q_{2} =298\: \: J / s\\  \\   \:  \:  \:  \sf \:  \bf \: \: P= Q_{2} -  Q_{1}  \\   \tt \: \:  \:  \:  \:  \:  \:  \:  \:  \:=298 - 263  \\ \\  \:  \:  \:  \tt \:  \:  \:  \:  \:  \:  \rightarrow \:  \fbox{ \blue{35 \: watt}}

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2 years ago
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