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Murljashka [212]
3 years ago
12

A small candle is 34 cm from a concave mirror having a radius of curvature of 22 cm. Where will the image of the candle be locat

ed? Follow the sign conventions.
Physics
1 answer:
xxTIMURxx [149]3 years ago
5 0

Answer:

16.26 cm in front of the mirror

Explanation:

Using,

1/f = 1/u+1/v....................... Equation 1

Where f = focal focal length of the concave mirror, u = object distance, v = image distance.

make v the subject of the equation

v = fu/(u-f)................... Equation 2

Note: The focal length of a concave mirror is positive

Using the real- is- positive convention

Given: f = 22/2 = 11 cm, u = 34 cm.

Substitute into equation 2

v = (34×11)/(34-11)

v = 374/23

v = 16.26 cm.

The image will be formed 16.26 cm in front of the mirror.

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The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

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The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

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$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

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Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the end
yawa3891 [41]

Correct question:

Manny walked a total of 3 miles. The reference point used to calculate the total distance that he walked was the same as the ending point. Which describes where Manny most likely walked?

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