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zubka84 [21]
4 years ago
7

Name and briefly describe the two types of interference

Physics
1 answer:
Annette [7]4 years ago
6 0

The two types of wave interference are constructive and destructive.

These both describe what happens when waves combine

<u>Constructive interference</u> is when 2 waves combine to form a wave with a larger amplitude, but this is only if both waves are both positive or both negative.

<u>Destructive interference</u> is when the two waves are opposite, one is negative, one is positive. They subtract forming the combined wave that has a lower amplitude.

I hope that helps u out!! :)


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How much power will it take to move a 10 kg mass at an acceleration of 2 m/s² a distance of 10 meters in 5 seconds?
DedPeter [7]

Answer:

100 Watts

Explanation:

<u>These equations are needed to work out the answer:</u>

  • power= work done/ time taken
  • work done= force* distance
  • force= mass* acceleration

force: 10 kg* 2m/s= 20

work done: 20* 10m=200

power: 200/2=100

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Maxwell shoots a rubber band at his friend Jimmy. Which type of energy is converted into kinetic energy?
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A stretched rubber band is storing <em>elastic potential energy. (A)</em>

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This type of light production is an example of _______ energy being transformed into light energy.
Anton [14]

Electric energy??? there isn't much information from the question but I can infer it's electricity.

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Por favor ayuda que ejercicio debo hacer para enflacar en una semana xd
Serggg [28]

Answer:

Un excelente ejercicio para perder barriga rápidamente es correr, ya que el cuerpo gasta una mayor cantidad de calorías en un corto período de tiempo, en tan sólo 30 minutos corriendo se queman unas 300 calorías.

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8 0
3 years ago
6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
1 year ago
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