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3 years ago

Tides are a result of

Physics

2 answers:

Virty [35]3 years ago

6 0

The answer should be D but if its not then im sorrry, idk

Mrrafil [7]3 years ago

6 0

Tides are a result of <u>"the gravitational attraction of the Sun and Moon".</u>

An especially high tide (Spring tide) happens when the sun and moon are in line and both pull in a similar way.

Tides may appear to be straightforward at first glance, yet the intricate details of tides jumbled extraordinary logical masterminds for a considerable length of time they even driven Galileo astray into a bunk theory.

Today individuals realize that the gravitational pulls between the earth, moon and sun manage the tides. The moon, however, impacts tides the most.

The moon's gravitational pull on the earth is solid enough to pull the seas into bulge. On the off chance that no different powers were impacting everything, shores would encounter one high tide a day as the earth pivoted on its hub and coasts kept running into the seas' bulge confronting the moon.

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Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1

harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation $E_p=h f$ , where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of $E_N=N h f$ .

We also know that frequency and wavelength are related by $f=\frac{c}{\lambda}$ , so we have $E_N=\frac{N h c}{\lambda}$ , where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

$N=\frac{\lambda E_N}{h c}$

We need to know then how much energy do we have to calculate N. The equation of power is $P=E/t$ , so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be $E_N$ .

Putting this paragraph in equations:

$E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J$ .

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

$N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}$

3 0

3 years ago

We can determine the velocity of a wave when given the frequency and the

Kobotan [32]

Hello.

The answer is: D. wavelength

This is correct because frequency x wavelength = speed

Have a nice day

3 0

3 years ago

Read 2 more answers

Describe The energy conversions that occur when coal is formed, and then burned to produce thermal energy.

svlad2 [7]

Answer:

When energy contained in coal is turned into heat, and then into electrical energy. As boiling water heated by the burning coal is cooled, steam forges from these cone-shaped cooling towers.

8 0

3 years ago

A box is held at rest by two ropes that form 30° angles with the vertical. The tension T in either rope is 42 N. What is the wei

iragen [17]

<h3><u>Answer;</u></h3>

= 73 N

<h3><u>Explanation</u>;</h3>

Using the formula

2 T cos(30°) = w

Where; T is the tension on each string, while w is the weight of the box given by mg

Therefore;

W = 2Tcos 30°

= 2 × 42 cos 30°

= 84 cos 30°

= 72.74

7 0

4 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$ ?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

Again, using the quadratic ecuation:

$t_2=24.49s$

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

$t_T=t_1+t_2=34.49s$ .

6 0

4 years ago