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tankabanditka [31]
2 years ago
6

A. Label the following properties on the wave below: rarefaction, wavelength, and compression. (Image attached)

Physics
2 answers:
DedPeter [7]2 years ago
0 0

PART a)

Rarefraction : It is the region of lower density where particles move away from each other

Wavelength : distance between two consecutive compression or rarefaction is known as wavelength

Compression : The region of higher density or the region where particles comes closer to each other is compression

Part b)

Here it is given that

f = 2100 Hz

\lambda = 3.0 m

So here we know that

wave speed = frequency \times wavelength

[tex]v = 2100(3) = 6300 m/s

PART c)

As the person drop the stone in the water it creates waves on the surface of water

These waves can move the boat up and down at its own position and it can not displace the boat so it will not move to the shore

PART D)

When stone is dropped in the water it creates waves. these waves when travel through the water and disturb the boat

Due to this disturbance boat starts moving up and down at its own position and it will not displace from its position.

So by this type of waves it will not move towards the shore

Alecsey [184]2 years ago
0 0

a) Rarefactions are in the green frames, compressions in the purple. Wavelength is a distance between same points of the two corresponding rarefactions/compressions.

b) v=wavelength times frequency=3*2100= 6300 m/s.

c) Bad idea that most likely won't work. Waves on water move only in vertical direction. But maybe a high wave will lift the boat and rescue it.

d) Ah, he needs wants the boat reach the shore, I thought it stuck and the person just wants to make it float. Yes, that's not the best idea, see explanation above

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Answer:

The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

     v  = \sqrt{\frac{G  M}{r^2} }

=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

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3 years ago
A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D
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Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

v² = u² + 2a s

vf² =  vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M  vi² + FD) / 1/2 M  vi²

RK = 1 + FD / 2 M  vi²

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Answer:

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Explanation:

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Two pieces of amber are hung from threads. Piece A is charged by rubbing piece A with fur. Piece B is charged by rubbing piece B
PIT_PIT [208]

Answer:

ieces A and B must also have the same type of charges

Explanation:

In electrostatics, charges of the same sign repel and charges of different signs attract.

If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.

Also part A and C repel each other, therefore they have the same type of charge.

If we use the transitive property of mathematics, pieces A and B must also have the same type of charges

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A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
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Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

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F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

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