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Tresset [83]
3 years ago
10

Part II # 1 A mass on a string of unknown length oscillates as a pendulum with a period of 4 sec. What is the period if: (Parts

(1-4 are independent questions, each referring to the initial situation.) 1) The mass is doubled? 2) The string length is doubled? 3) The string length is halved? 4) The amplitude is halved? A. 1) No Change (4 sec) 2) 2.8 sec 3) 5.7 sec 4) No Change (4 sec) B. 1) 5.7 sec 2) No Change (4 sec) 3) 2.8 sec 4) No Change (4 sec) C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec) D. 1) 5.7 sec 2) No Change (4 sec) 3) No Change (4 sec) 4) 2.8 sec
Physics
1 answer:
Mrac [35]3 years ago
6 0

Answer:

C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec)

Explanation:

Given that:

Period (T) = 4 s

1) If the mass is doubled.

The period of a pendulum is given by the formula:

T=2\pi\sqrt{\frac{L}{g} } where L is the length and g is the acceleration due to gravity.

From the formula, the period does not depend on the mass of the spring therefore if the mass is doubled the period does not change.

2) The string length is doubled

Given that:

T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }

if the length is doubled, the new spring length is 2L. Therefore the new period (T1) is given as:

T_1=2\pi\sqrt{\frac{2L}{g} }=\sqrt{2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{2}*4=5.7\ sec

3) The string length is halved

Given that:

T=2\pi\sqrt{\frac{L}{g} }, but\ T =4s\\4=2\pi\sqrt{\frac{L}{g} }

if the length is halved, the new spring length is L/2. Therefore the new period (T1) is given as:

T_1=2\pi\sqrt{\frac{L}{2g} }=\sqrt{1/2} (2\pi\sqrt{\frac{L}{g} })=\sqrt{1/2}*4=2.8\ sec

4) The amplitude is halved

From the formula, the period does not depend on the amplitude therefore if the amplitude is halved the period does not change.

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Answer:

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Explanation:

1. Determination of the force of attraction.

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Distance apart (r) = 500 m

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The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

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Answer:

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Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

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Density = mass / volume

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Cross multiply

Mass of brick in water = 1 × 1900

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Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

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Next, we shall determine the weight in water. This can be obtained as follow:

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