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3 years ago

Are waves matter or not?

Physics

1 answer:

STatiana [176]3 years ago

5 0

Answer:

A Wave Transports Energy and Not Matter

When a wave is present in a medium (that is, when there is a disturbance moving through a medium), the individual particles of the medium are only temporarily displaced from their rest position.

Explanation:

You might be interested in

Question 1

sineoko [7]

Velocity = displacement / time
Displacement = 2.3 km
Time = 5.78 mins
Velocity = 2.3/5.78
Velocity = 0.398 to the nearest 3 significant figures

4 0

3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago

What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87 m?

7nadin3 [17]

V = 174 m/s
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8 0

3 years ago

Read 2 more answers

Three resistors having resistances of LaTeX: 4.0\;\Omega4.0 Ω, LaTeX: 6.0\;\Omega6.0 Ω, and LaTeX: 10.0\;\Omega10.0 Ω are connec

aalyn [17]

Answer:

Explanation:

Given that three resistor of resistance

R1 = 4 Ω

R2 = 6 Ω

R3 = 10 Ω

The resistor are connected in parallel, the equivalent resistance is

1 / Req = 1 / R1 + 1 / R2 + 1 / R3

Rearranging this we have

Req = R1•R2•R3 / (R2•R3 + R1•R3 + R1•R2)

Req = 4 × 6 × 10 / (6×10 + 4 × 10 + 4 × 6)

Req = 240 / (60 + 40 + 24)

Req = 240 / 124

Req = 1.94 Ω

The parallel connection is connected in series with a battery and a resistor.

Resistor resistance is

r = 2 Ω

Supply voltage of the battery is 12V

V = 12V

Let find the current flowing in the circuit.

V = I(Req + r)

I = V / (Req + r)

I = 12 / (1.94 + 2)

I = 12 / 3.94

I = 3.05A

So, this is the current flowing in the circuit and It is the same current that will be shared by the parallel resistance.

We can calculated the voltage across the parallel resistance

From ohms las

V = iR

V = I × Req

V = 3.05 × 1.94

V = 5.92 V.

Then, this the voltage across each parallel resistor.

Then, to know the current in the 10ohms resistance

V = iR

I = V / R3.

I = 5.92 / 10

I = 0.592 A.

The current in the 10 ohms resistor is 0.59A

5 0

3 years ago

A skater of mass 60 kg has an initial velocity of 12 m/s. He slides on ice where the frictional force is 36 N. How far will the

Alexus [3.1K]

Answer:

d = 120 [m]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. Where the energy in the final state (when the skater stops) is equal to the sum of the mechanical energy in the initial state plus the work done on the skater in the initial state.

The mechanical energy is equal to the sum of the potential energy plus the kinetic energy. As the track is horizontal there is no unevenness, in this way, there is no potential energy.

E₁ + W₁₋₂ = E₂

where:

E₁ = mechanical energy in the initial state [J] (units of Joules)

W₁₋₂ = work done between the states 1 and 2 [J]

E₂ = mechanical energy in the final state = 0

E₁ = Ek = kinetic energy [J]

E₁ = 0.5*m*v²

where:

m = mass = 60 [kg]

v = initial velocity = 12 [m/s]

Now, the work done is given by the product of the friction force by the distance. In this case, the work is negative because the friction force is acting in opposite direction to the movement of the skater.

W₁₋₂ = -f*d

where:

f = friction force = 36 [N]

d = distance [m]

Now we have:

0.5*m*v² - (f*d) = 0

0.5*60*(12)² - (36*d) = 0

4320 = 36*d

d = 120 [m]

7 0

3 years ago