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ohaa [14]
3 years ago
14

A wave passes along the surface of the water in a ripple tank. Describe the motion of a molecule on the surface as the water pas

ses.
Physics
1 answer:
laila [671]3 years ago
8 0

Answer:

Explanation:

Water waves are generally a transverse wave which do not cause permanent displacement of molecules of the medium. Transverse waves are waves in which the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium.

As the wave propagates from one point to another on the surface of water transferring energy, a molecule of water on its surface vibrates upwards and downwards. Its motion is perpendicular to the direction of propagation of the wave. After the vibration, it comes back to its initial position.

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A. The sum of all the forces acting on an object.

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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
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Answer:

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