All categories

3 years ago

What rock type would most likely be found in the location marked by the black rectangle in the diagram below?

Physics

2 answers:

pychu [463]3 years ago

5 0

There is no diagram below so I can't answer the question

Brums [2.3K]3 years ago

5 0

Just took the assessment its Metamorphic.

You might be interested in

Why can a star have several different habitable zones

Rzqust [24]

Because they are placed in different habitable zones

8 0

3 years ago

PLEASE I NEED HELP I AM REALLY STUCK!!!

AlekseyPX

Answer: Force = Mass X Acceleration

F = 5 x 2

F = 10 N

8 0

3 years ago

Name two factors that can affect the function of an enzyme

myrzilka [38]

Answer:

1. pH

2. Temperature

Hope this helps

3 0

3 years ago

Which statement is correct?

Svetach [21]

Out of the choices given, the statement about how light travels is "<span>Light can travel in a vacuum, and it travels faster if the light source is moving."</span>

7 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