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3 years ago

6

something that is figurative place where a person locates the source of responsibility in his or her life.

1 answer:

ahrayia [7]3 years ago

4 0

What is your choses .becuse it mit be a place

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The main source of information used by astronomers to learn about objects in space is.

Dmitriy789 [7]

Electromagnetic waves

4 0

2 years ago

Which of the following is a characteristic of digital data?

Westkost [7]

Answer:

duplication being slightly more expensive.

Explanation:

3 0

3 years ago

A lawyer drives from her home, located 1 mile east and 8 miles north of the town courthouse, to her office, located 4 miles w

givi [52]

Answer:

d = 13 miles

Explanation:

Lets say the position of court house is origin in this case

her office is located at 4 miles west and 4 miles south of court house

so here we have coordinate of the office with respect to court house is given as

$r_1 = (-4\hat i - 4\hat j)$

now the position of her home is located at 1 miles east and 8 miles north of the court house

so the coordinates of her home is given as

$r_2 = (1\hat i + 8 \hat j)$

now the change in the position is given as the distance between office and home

$d = r_2 - r_1$

$d = 5 \hat i + 12 \hat j$

$d = \sqrt{5^2 + 12^2} = 13 miles$

4 0

4 years ago

The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the si

JulijaS [17]

Answer:

0.75

Explanation:

$refractive \: index \: = \frac{3.0 \times {10}^{2} }{2.0 \times {10}^{2} }$

= 1.5

$refractive \: index = \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }$

$1.5 = \ \frac{ \sin(i) }{ \sin(30) }$

1.5 × ½ = sin(i)

$\sin(i) = 0.75$

5 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago