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VashaNatasha [74]
3 years ago
11

What is a transverse wave​

Physics
2 answers:
Anvisha [2.4K]3 years ago
6 0

Answer:

In Physics, a transverse wave is a moving wave whose oscillations are perpendicular to the direction of the wave. A simple demonstration of the wave can be created on a horizontal length of the string by securing one end of the string and moving the other up and down. Light is another example of a transverse wave, where the oscillations are electric and magnetic fields that are at right angles to the ideal light rays that describe the direction of propagation.

Explanation:

In Physics, a transverse wave is a moving wave whose oscillations are perpendicular to the direction of the wave. A simple demonstration of the wave can be created on a horizontal length of the string by securing one end of the string and moving the other up and down. Light is another example of a transverse wave, where the oscillations are electric and magnetic fields that are at right angles to the ideal light rays that describe the direction of propagation.

Transverse waves commonly occur in elastic solids, oscillations, in this case, are the displacement of solid particles from their relaxed position, in the direction perpendicular to the propagation of the wave. Since these displacements correspond to local shear deformation of the material, the transverse waves of this form are known as a shear wave. In seismology, shear waves are also known as secondary waves or S-waves.

Examples of Transverse waves

Some examples of transverse waves are listed below:

The ripples on the surface of the water

Electromagnetic waves

Stadium or a human wave

Ocean Waves

The secondary waves of an earthquake

(I hope this was helpful) >;D

tankabanditka [31]3 years ago
5 0

Answer:

a wave vibrating at right angles to the direction of its propagation.

Explanation:

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Calcula el peso aparente de una bola de aluminio de 50 cm3, cuando se encuentra totalmente sumergida en alcohol. Datos: la densi
artcher [175]

Answer:

W_apparent = 93.1 kg

Explanation:

The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.

            W_apparent = W - B

The push is given by the expression of Archimeas

            B = ρ_fluide g V

            ρ_al = m / V

            m = ρ_al V

we substitute

            W_apparent = ρ_al V g - ρ_fluide g V

            W_apparent = g V (ρ_al - ρ_fluide)

       

we calculate

           W_apparent = 980 50 (2.7 - 0.8)

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7 0
3 years ago
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

  • Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
  • This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
  • Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
  • According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

       a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)

  • Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

       v_{f} ^{2}  -v_{o} ^{2} = 2* a* \Delta x (2)

  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

       \Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)

b)

  • We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
  • Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

7 0
3 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

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Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.
3 0
3 years ago
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