Ask question

Ask question

All categories

3 years ago

9

70 POINTS AND BRAINLIEST PLEASE HELP!!!The Moon orbits Earth. This orbit causes the Moon to look different over the course of ab

out a month. These differences are called phases. One phase is called a full moon. Look closely at the table to the right. What can you conclude about a full moon? A. A. A full moon only occurs in the summer months. B. B. A full moon always occurs after a first quarter moon. C. C. A full moon always occurs on the 5th day of the month. D. D. A full moon always occurs before the first quarter moon.

1 answer:

Rudik [331]3 years ago

8 0

Answer:

It always acurs after a 1st quarter

do you have photo?

Explanation:

You might be interested in

Question 20 Unsaved The graph below represents changes in water. What process is occurring during section "B" of the graph? Ques

FinnZ [79.3K]

melting ... happpens at constant temp

4 0

3 years ago

Read 2 more answers

Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m

Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

5 0

3 years ago

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

umka2103 [35]

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

mass of raindrops, $m=1.8\times 10^{-6}\ kg$

charge on the raindrops, $q=+21\times 10^{-12}\ C$

horizontal distance between the raindrops, $r=0.0044\ m$

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

<em>Now force:</em>

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

7 0

3 years ago

Read 2 more answers

What are potential impacts of pollution on a watershed? Check all that apply.

Studentka2010 [4]

Answer:

contaminated drinking water

deaths of sea creatures that are used as a food source

limits to potential economic activities such as a fishing

Explanation:

A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.

Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.

3 0

3 years ago

Read 2 more answers

Approximately where is it currently high tide on Earth? Group of answer choices wherever it is currently noon anywhere that ocea

Wewaii [24]

Answer:

Option D, only on the portion of the Earth facing directly toward the Moon

Explanation:

Tides are caused by the gravitational pull of moon. The part of earth that faces the moon experiences the highest gravitational force and hence the high tides will occur in this regions only. The regions that do not faces the moon experiences low tides. It is the gravity of moon that attracts the ocean water towards itself.

Hence, Option D is correct

7 0

2 years ago