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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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I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

3 years ago

The speed of light in a vacuum is 2.99x10^8 m/s. calculate its speed in miles per hour.

34kurt

You'll never get the correct answer without the correct conversion factor. Note carefully that you have no decimal. It should be
1 km = 0.6214 miles 
1000 m = 1 km 
60 seconds = 1 minute 
60 minutes = 1 hour. 
2.998E8 m/s x (1 km/1000m) x (0.6214 miles/km) x (60 sec/min) x (60 min/hr) = ?

6 0

3 years ago

In general, light waves travel _____ sound waves.

larisa [96]

A. more quickly. example lightning (light) comes first in a storm. then thunder (sound) comes after

5 0

3 years ago

Read 2 more answers

What volume will be occupied by 20g of Copper if the density of Copper is 9.1g/ml?

Eva8 [605]

Answer:

The volume of copper is 2.198 ml

Explanation:

Given;

mass of copper, m = 20 g

density of copper, ρ = 9.1 g/ml

Density is given by;

Density = mass / volume

Volume = mass / density

Volume = (20 g) / (9.1 g/ml)

Volume = 2.198 ml

Therefore, the volume of copper is 2.198 ml

3 0

3 years ago

Which voltage is expected to have higher accuracy 20v or 200v?

gayaneshka [121]

Answer:

20v

Explanation:

Smarts

4 0

3 years ago