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tamaranim1 [39]
3 years ago
15

Camera lenses are described in terms of

Physics
1 answer:
Lesechka [4]3 years ago
6 0

Answer:

The position of the image from the camera lens is approximately 121.1 mm

Explanation:

The given parameters of the lens are;

The specification of the camera lens = 120 mm

Therefore, the focal length of the camera lens, f = 120 mm = 0.12 m

The distance of the object from the camera, d_o = 13 m

The lens equation for finding the position of the image is given as follows;

\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}

Where;

d_i = The position of the image from the camera lens

Therefore, by plugging in the known values, we have;

\dfrac{1}{13} = \dfrac{1}{0.12} + \dfrac{1}{d_i}

\dfrac{1}{d_i}  = \dfrac{1}{0.12} -  \dfrac{1}{13} = \dfrac{13 - 0.12}{0.12 \times 13} = \dfrac{12.88}{1.56} = \dfrac{322}{39}

\therefore d_i = \dfrac{39}{322}  \approx  0.1211

The position of the image from the camera lens, d_i = 0.1211 m = 121.1 mm

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If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
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The third force must balance them, in order to have a net force of zero on the object.

The resultant of the two forces F1 and F2 is
F_{12} =  \sqrt{F_1^2+F_2^2}= \sqrt{(10 N)^2+(10 N)^2}= \sqrt{200}=14.1 N
with direction at 45^{\circ} north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be 45^{\circ} south-east.
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Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

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Using formula of angular velocity

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\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi

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A diver 40 m deep in 10 degrees C fresh water exhales a 1.5 cm diameter bubble.
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