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Vladimir79 [104]
4 years ago
14

A small candle is 38 cm from a concave mirror having a radius of curvature of 24 cm. (a) what is the focal length of the mirror?

(
Physics
1 answer:
Finger [1]4 years ago
8 0
For a concave mirror, the radius of curvature is twice the focal length of the mirror:
r=2f
where f, for a concave mirror, is taken to be positive.

Re-arranging the formula we get:
f= \frac{r}{2}
and since the radius of curvature of the mirror in the problem is 24 cm, the focal length is
f= \frac{24 cm}{2} = 12 cm

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Suppose you walk 11 m in a direction exactly 24° south of west then you walk 21 m in a direction exactly 39° west of north. 1) H
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(1) 42.94 m

(2) 16.02^\circ

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Let us first write the two displacements in the vector form:

\vec{AB} = (-11\cos 24^\circ\ \hat{i}-11\sin 24^\circ\ \hat{j})\ m =(-10.05\ \hat{i}-4.47\ \hat{j})\ m\\\vec{BC} = (-21\sin 39^\circ\ \hat{i}+21\cos 39^\circ\ \hat{j})\ m =(-31.22\ \hat{i}+16.32\ \hat{j})\ m

Now, the vector sum of both these vectors will give us displacement vector from point A to point C.

\vec{AC}=\vec{AB}+\vec{BC}\\\Rightarrow \vec{AC}=(-10.05\ \hat{i}-4.47\ \hat{j})\ m+(-31.22\ \hat{i}+16.32\ \hat{j})\ m\\\Rightarrow \vec{AC}=(-41.25\ \hat{i}+11.85\ \hat{j})\ m

Part (1):

the magnitude of the shortest displacement from the starting point A to point the final position C is given by:

AC=\sqrt{(-41.25)^2+(11.85)^2}\ m= 42.94\ m

Part (2):

As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.

The angle of the shortest line connecting the starting point and the final position measured north of west is given by:

\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ

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