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exis [7]
2 years ago
10

A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Which of the fo

llowing statements are true when the system undergoes small oscillations?
Check all that apply.

A. The period is inversely proportional to the suspended mass.
B. The period is proportional to the square root of the length of the wire.
C. The period is independent of the suspended mass.
D. The period is proportional to the suspended mass.
E. The period is independent of the length of the wire.
F. The period is inversely proportional to the length of the wire.
Physics
1 answer:
jarptica [38.1K]2 years ago
4 0

Answer:

- The period is independent of the suspended mass.

- The period is proportional to the square root of the length of the wire.

Explanation:

A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Which of the following statements are true when the system undergoes small oscillations?

Check all that apply.

A. The period is inversely proportional to the suspended mass.

B. The period is proportional to the square root of the length of the wire.

C. The period is independent of the suspended mass.

D. The period is proportional to the suspended mass.

E. The period is independent of the length of the wire.

F. The period is inversely proportional to the length of the wire.

Simple harmonic motion is periodic motion under the action of a restoring force that is directly proportional to the displacement from equilibrium

from the relation of period T

T=2\pi \sqrt{l/g}

from the above formula , it can be concluded that

C. The period is independent of the suspended mass

B.The period is proportional to the square root of the length of the wire.

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Answer

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                           I_s =\frac{ V}{R_T}

                            I_s  = \frac{V_0}{R_1 +R_2} ---(1)

The mathematical representation of the two resistors connected in parallel  is

                    R_T = \frac{1}{R_1} +\frac{1}{R_2}

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From the question I_p =10I_s

          =>                 I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} }  = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)

     Dividing equation 2 with equation 1

       =>                 \frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}

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                            10 = \frac{(1-r)^2}{r}

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=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0

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        a = 1  b = -8 c =1  

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                               r= \frac{8+ \sqrt{60} }{2}  \ or \  r = \frac{8 - \sqrt{60} }{2}

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Now  r =  0.127 because it is the least value among the obtained values

                               

                                   

                             

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Answer:

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Temperature T_{1}= 20.0°C = 293\ K

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Hence, The volume at mountains is 2.766 L.

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