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steposvetlana [31]
2 years ago
15

A 9.0 cm object is 3.0 cm from a lens, which has a focal length of –12.0 cm.

Physics
2 answers:
artcher [175]2 years ago
7 0
The answers are -2.4 for the distance. 7.2 for the height. Concave for the type
g100num [7]2 years ago
3 0

Answer :

Image distance = -2.4 cm

Height of image = 7.2 cm

Concave lens

Explanation :

Given that,

Height of the object, h = 9 cm

Object distance, u = 3 cm

Focal length, f = -12 cm

Using mirror's formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{(-12)}-\dfrac{1}{(3)}

v = -2.4 cm

Image distance is 2.4 cm from the lens.

Magnification, m =\dfrac{-v}{u}=\dfrac{h'}{h}

h'=\dfrac{-vh}{u}

h' = 7.2 cm

The height of the image is 7.2 cm. This shows that the image is upright.

The lens used is concave lens.

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Explanation:

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Put the value into the formula

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Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

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