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Helen [10]
3 years ago
11

You drop a rock into a pond, and water waves spread out in circles. (a) The waves carry water outward, away from where the rock

hit. That moving water carries energy outward. (b) The waves only make the water move up and down. No energy is carried outward from where the rock hit. (c) The waves only make the water move up and down, but the waves do carry energy outward, away from where the rock hit.
Physics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:

Explanation:

wave motion is defined as the propagation of the disturbance due to continous vibration of the molecules of the medium. Waves oduced due to disturbance carry only energy but not the matter.

The waves move water up and down and carry the energy.

Therefoe, the statement "(a) The waves carry water outward away from where the rock hit. The moving water carries energy outward"is incorrect.

The waves move water up and down and carry the energy, but not matter.

Therefore, the statement "(c) The waves ony make the water move u and down. No energy is carried outward from where the rock hit" is incorrect.

When a rock is droed ito a pd, a disturbance starts from the point where the stone hits the water. The waves produced carry the energy from the starting point toother locatons over a large distance which is not possible by the single molecules of water.

Therefore, the statement "(c) The waves only make the water move up and down, but the waves do carry energy outward away from where the rock hit" is correct.

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3 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
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Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

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What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k
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Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

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