What measures the distance between two consecutive crests of a wave?

Answers

Answer 1

Answer: The distance between two successive crests or two successive troughs is the wavelength for a transverse wave. The height of the crest from the undisturbed water position or the depth of the trough from the undisturbed water position is the amplitude of the wave.

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Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far from it would you place a sphere representing the earth? (Center to center distance please.) The distance from the center of the sun to the center of the earth is 1.496×10111.496×1011 m and the radius of the sun is 6.96×1086.96×108 m.

Answers

In this exercise we have to use the knowledge in distance, in this way we will find that the proportional distance found is:

$d = 0.645 m$

So from the information given in the text we find that:

The distance from the center of the sun to the center of the earth is $1.496*10^(11) \ m$

The radius of the sun is $6.96*10^(8)m$

We need to assume a radius for the ball bearing, so suppose that the radius is $3 mm$

First, we need to find in what way or manner often the radius of the brightest star exist considerable respect to the range of the ball significance, that exist given apiece following equating:

$(r_a)/(r_b)= \frac{6.96*10^8}{3*10{-3}} =2.32*10^(11)$

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth:

$d_(s) = (d_(e))/(r_(s)/r_(b)) = (1.496 \cdot 10^(11) m)/(2.32\cdot 10^(11)) = 0.645 m$

See more about distance at brainly.com/question/989117

Answer:

d = 0.645 m(assuming a radius of the ball bearing of 3 mm)

Explanation:

The given information is:

The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = $d_(e)$
The radius of the sun is 6.96x10⁸m = $r_(s)$

We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = $r_(b)$ .

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

$(r_(s))/(r_(b)) = (6.96\cdot 10^(8)m)/(3\cdot 10^(-3)m) = 2.32\cdot 10^(11)$

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, $d_(s)$ :

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!

What is the mass of a pure platinum disk with a volume of 113 cm3? The density of platinum is 21.4 g/cm3.Give your answer in grams and kilograms.

Answers

The mass of the platinum if the volume is 113 cm³ and density is 21.4 g / cm³ is 2.418 kilograms or 2 kilograms and 418 grams.

What is density?

The density is the mass of a material substance per unit volume. d = M/V, where d is density, M is mass, and V is volume, is the formula for density. Grams per cubic centimeter are a typical unit of measurement for density.

As an illustration, the density of Earth is 5.51 grams per cubic centimeter, whereas the density of water is 1 gram per cubic centimeter.

Given:

The volume of the platinum disk, V = 113 cm³,

The density of platinum, d = 21.4 g / cm³

Calculate the mass using density and mass relation as shown below,

density = Mass of the platinum / Volume of the platinum

21.4 = Mass of the platinum / 113

Mass of the platinum = 21.4 × 113

Mass of the platinum = 2418.2 g

As we know 1 kg = 1000 grams,

Mass of the platinum = 2418.2 / 1000

Mass of the platinum = 2.418 kg

Thus, the mass of the platinum is 2.418 kg.

To know more about Density:

brainly.com/question/6329108

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Multiply the density with the volume. You get 2418.2 grams or 2.4182 kilograms :)

Humpty Dumpty (12 kg) sat on a wall (2 m high). What was his potential energy, before his great fall? A. 235.2 J B. 235.2 N C. 24 J D. 246.1 W

Answers

I think it would be B. P.E.=mgh, the mass is 12g, gravitational force is 9.8, and the wall is 2 meters high, so multiply it all and get 235.2.

It would be 24J because there's no energy.

How many electrons will there be in 5 coulombs of charge? (the charge of one electron is 1.6 x 10^-19 Coulombs)
thanks!

Answers

There's no physics or electronics to this question.
It's just arithmetic.

The last part of the question TELLS you how many 'coulombs per electron'.

If you just flip that fraction (divide ' 1 ' by it, take its reciprocal), then
you'll have 'electrons per coulomb', and 5 of those will answer the question.

I got 31,250,000,000,000,000,000 . I could be wrong. You should check it.

During which period did the first large herbivores and carnivores appear?

Answers

The first large herbivores and carnivores appeared during the Mesozoic Era, which is divided into three periods-- Triassic, Jurassic and Cretaceous. The herbivores and and carnivores increased in size during the Jurassic Period. And some of the largest dinosaurs emerged during this period as well.

A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3 kg moving in the negative X-Direction with a speed of 1 m/s . During collision a loud sound is heard and they both start moving together . the sound energy cannot be greater than :A) 12 J
B) 14 J
C) 15 J
D) 17.5 J

Answers

$m_1=2 \n m_2=3 \n v_1=4 \n v_2=1 \n v\text{ =speed after collision (to be determined)}.$

The momentul of the system preserves:

$m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=(m_1v_1-m_2v_2)/(m_1+m_2).$

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with $E$ .

Now, we write the energy conservation law:

$(m_1v_1^2)/(2)+(m_2v^2_2)/(2)=((m_1+m_2)v^2)/(2)+E$

From the above equation, you find $E$ , and then conclude that the sound energy can certainly not be greater than this.

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