All categories

3 years ago

Describe the interactions with the Sun, Earth and Moon system that result in Tides.

Physics

1 answer:

bazaltina [42]3 years ago

3 0

Answer:The sun, earth, and moon are held together by gravity, and they interact in lots of ways.

The tides are another interaction in the sun-earth-moon system. The tides happen because the moon and sun pull on the oceans, causing them to rise and fall each day. The moon has a bigger effect than the sun because it is closer.

You might be interested in

If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di

xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

5 0

3 years ago

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce

jasenka [17]

Answer:

$\lambda= 506.25 nm$

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

$y=\frac{m \lambda D}{a}$

Where:

$y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1$

Solving for λ:

$\lambda=\frac{y*a}{mD}$

Replacing the data provided by the problem:

$\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm$

7 0

3 years ago

Which layer of the earth includes the ozone layer

Rus_ich [418]

Troposphere is the answer

7 0

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$