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8 days ago

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A 15.0 cm object is 12.0 cm from a concave mirror that has a focal length of 4.8 cm. Its image is 8.0 cm in front of the mirror.

What is the approximate height of the image produced by the mirror? –4 cm –10 cm 10 cm 4 cm

2 answers:

kicyunya [2.8K]8 days ago

7 0

The correct answer is -10.

Ostrovityanka [2.7K]8 days ago

3 0

To explain: The height of the object, h, equals 15 cm. The object distance, u, is -12 cm (this is negative for a concave mirror). The focal length of the concave mirror, f, is -4.8 cm. The image distance, v, equals -8 cm. The height of the image, h' is unknown. We calculate magnification as follows: h' = -10 cm. Hence, the image height is 10 cm and is upside down, confirming that option (B) '-10 cm' is correct.

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For an oscillating spring-mass system, the time period is expressed as:

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b) When the stiffness constant is quadrupled, holding other factors constant:

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Thus, the new time period:

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

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c) When both mass and stiffness constant are quadrupled:

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d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

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