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3 years ago

Which waves move by replacing one particle with another

2 answers:

8 0

Question : Which waves move by replacing one particle with another

Answer : Sound waves

Why : Well because sound waves only work when music or a clip is playing. If you stop it stops completely and nothing else. If you replace one particle on a sound wave the sound is going to change depending on what type of particle you changed!

-procklown

sweet-ann [11.9K]3 years ago

4 0

It should be Sound waves

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Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of res

Nesterboy [21]

Answer:

"Narrow the focus of research question"

Explanation:

O Narrow the focus of research question

This is good! You can still use your question, but focus in on something so you have a proper research project.

O Add another research question

Would adding another question to an already broad question help? No.

O Use the very first source you find for your project

If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question

O Change the scope of your project

You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

3 0

2 years ago

1000 kg of water initially at 6 m/s runs through a hydro-generator. If the water leaves the generator at velocity of 4 m/s, and

r-ruslan [8.4K]

Answer:

8.00 kJ

Explanation:

The first thing is to determine what quantities are there.

the mass of water = 1 000 kg

initial velocity, u = 6 m/s

final velocity, v = 4 m/s

the generator is operating at 100 % efficiency, so there is no energy loss.

The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.

The kinetic energy is calculated as follows:

Ek = 1/2 mv²

= 1/2×(1 000)× (4)²

= 8 000 J/s

= 8.00 kJ Ans

4 0

4 years ago

• what is the typical distance between two adjacent pins on a 14-pin dual-in-line ic package?

muminat

A 14 pin dual-in-line IC package[14 DIL] is an integrated socket which is most popular form of IC package and has a wide range of application in digital electronics.

The 14-pin DIL has two pairs per side and each pair contains seven connecting pins.

The pairs of pins are arranged linearly one after another.The typical dimensions of width is 6.5 mm and the typical dimension of length is 18 mm.

we are asked to calculate the typical distance between two adjacent pins.

The typical distance between two adjacent pins is calculated as-

$Typical\ distance =\frac{dimensional\ length}{number\ of\ pins\ in\ each\ row}$

$=\frac{18 mm}{7}$

$= 2.5714 mm$ [ans]

7 0

3 years ago

Science<br> plz help<br> will give brainlist

dezoksy [38]

1. It cools and condenses
2. Dew point
3. Clouds
(I’m sorry that’s all I knew)

6 0

3 years ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

7 0

3 years ago