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Eddi Din [679]
3 years ago
14

Objects floating in the water, like buoys, only bob up and down when waves pass. Why do they not get pushed all the way to where

ver the wave goes
Physics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

Because as the waves propagates, the particles of the medium (molecules of water) vibrates perpendicularly (upward and downward) about their mean position and not in the direction of the waves.

Explanation:

A wave is a phenomena which causes a disturbance in a medium without any permanent deformation to the medium. Examples are; transverse wave and longitudinal wave. Waves transfer energy from one point in the medium to another.

The waves generated by water are transverse waves. Which are waves in which the vibrations of the particles of the medium is perpendicular to the direction of propagation of the waves.

Thus as the waves propagates, the molecules of water vibrates up and down and not along the direction of propagation of the waves. So that the floating objects do not get pushed in the direction of the waves every time.

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3 years ago
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What gives the gem amethyst it's purplish color?
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3 years ago
What happens when light enters water?
Svet_ta [14]
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8 0
3 years ago
A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor
STatiana [176]

Answer:

The magnitude of the second charge is \rm 1.062\times 10^{-7}\ C or \rm 0.1062\ \mu C.

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge q_o at a point in the electric field of another charge q is given by the product of the amount of charge q_o and electric potential at that point due to the charge q.

U = q_o\ V.

The electric potential at that point is given by

V = \dfrac{kq}{r}.

where k is the Coulomb's constant.

Therefore,

U=q_o\ \dfrac{kq}{r}.

Now, We have given two charges q_1 = +44\ \mu C = +44\times 10^{-6}\ C and q_2, whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

U_i = \dfrac{kq_1q_2}{\infty}=0.

When the coordinates of position of the two charges are

(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).

The distance between the two charges is given by

r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.

The electric potential energy of the charges in this configuration is given by

U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.

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3 years ago
A sample of an unknown liquid has a volume of 12.0 mL and a mass of 6 g. What is its density? Show your work or explain how you
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The answer is 6.12 L
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