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Nat2105 [25]
4 years ago
10

Please help me.

Physics
2 answers:
cricket20 [7]4 years ago
8 0

Answer:

go to khan acamedy

Explanation:

vesna_86 [32]4 years ago
4 0

A) the tension in the string.

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Suppose a convex mirror has a focal length of 120 cm. A candle sits directly in front of the mirror. If the image of that candle
Andru [333]
We can solve the problem by using the mirror equation:
\frac{1}{f} = \frac{1}{d_o}+ \frac{1}{d_i}
where
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

For the sign convention, the focal length is taken as negative for a convex mirror:
f=-120 cm
and the image is behind the mirror, so virtual, therefore its sign is negative as well:
d_i=-24 cm
putting the numbers in the mirror equation, we find the distance of the object from the mirror surface:
\frac{1}{d_o} = \frac{1}{f}- \frac{1}{d_i}= \frac{1}{-120 cm} - \frac{1}{-24 cm}= \frac{1}{30 cm}
So, the distance of the object from the mirror is d_o = 30 cm
8 0
3 years ago
Suppose the wavelength of the light is 450 nm . how much farther is it from the dot on the screen in the center of fringe e to t
lbvjy [14]

Since you are looking at a right fringe, the answer is all the time a multiple of the wavelength for the reason that the light makes constructive intrusion. If you are looking at the center (zeroth) fringe, the difference is (450)(0) = 0. For the first maximum, the difference is 450 nm. For the second, it is 450 x 2 = 900 nm. Basically, for the nth maxima, the difference is 450n. 

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3 years ago
The cycle is a process that returns to its beginning, but it does not repeat
madreJ [45]
False. It does repeat itself
5 0
3 years ago
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