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morpeh [17]
2 years ago
8

A sound wave traveling eastward through air

Physics
2 answers:
Sholpan [36]2 years ago
6 0

Answer : A sound wave traveling eastward through air causes the air molecules to vibrate east and west

Explanation :

We know that the sound wave is a longitudinal wave. The particles of medium travels in the direction of propagation of the wave. The space where the particles are closely spaced is called compression and the space where the particles are far apart is called rarefaction. Also, the distance between two consecutive compression and rarefaction is called the wavelength.

If the sound wave is travelling in eastward direction, then the air molecules will move in the direction of propagation of wave. They will vibrate either in east or west direction.  

So, the correct option is (1) " vibrate east and west ".    

Wewaii [24]2 years ago
5 0
A sound wave traveling eastward through air<span> causes the air molecules to Vibrate east and west. </span>
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Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
Westkost [7]

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

5 0
2 years ago
Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at
aliina [53]

Answer:

a) 8.99*10³ V  b) 4.5*10⁻² J c) 0 d) 0

Explanation:

a)

  • The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
  • For a point charge, it can be expressed as follows:

        V =\frac{k*q}{d}

  • As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
  • This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
  • In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at  any other corner, as follows:

        V = \frac{2*q*k}{d} = \frac{2*8.99e9N*m2/C2*4e-6C}{8m} =\\ \\ V= 8.99e3 V

  • The potential at point C is 8.99*10³ V

b)

  • The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

        W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J

  • The work needed is 0.045 J.

c)

  • If we replace one of the charges creating the potential at the point  C, by one of the same magnitude, but opposite sign, we will have the following equation:

       V = \frac{8.99e9N*m2/C2*(4e-6C)}{8m}  + (\frac{8.99e9N*m2/C2*(-4e-6C)}{8m}) = 0

  • This means that the potential due to both charges is 0, at point C.

d)

  • If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.
5 0
2 years ago
You are moving at a speed 2/3 c toward randy when randy shines a light toward you. at what speed do you see the light approachin
yarga [219]
I see the light moving exactly at speed equal to c.

In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
</span>
The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.
3 0
1 year ago
A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a
JulijaS [17]

All the weight of the wooden board is bear by the support located at the centre of the rod, and the other support which is located at the end, will have no reaction force, or 0 reaction force.

Therefore the reaction at the centre support is equal to the weight of the board, while the support at the end has 0 reaction force.

8 0
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Answer:

a) battery-->electrical current-->copper wire rotor -->magnet--> mechanical energy

Explanation:

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