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IgorC [24]
2 years ago
11

Si se aumenta la tentación en una cuerda, ¿que pasa con la velocidad de la onda?​

Physics
1 answer:
Leokris [45]2 years ago
4 0

Answer:

El aumento de tensión alarga la longitud de onda, reduce la amplitud, aumenta la frecuencia y, por lo tanto, aumenta la velocidad. (GOOGLE)

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A table tennis ball with a mass of 0.003 kg and a soccer ball with a mass of 0.43 kg or both Serta name motion at 16 M/S calcula
poizon [28]
Let us first know the given: Tennis ball has a mass of 0.003 kg, Soccer ball has a mass of 0.43 kg. Having the same velocity at 16 m/s. First the equation for momentum is P=MV P=Momentum M=Mass V=Velocity. Now let us have the solution for the momentum of tennis ball. Pt=0.003 x 16 m/s= (    kg-m/s ) I use the subscript "t" for tennis.  Momentum of Soccer ball Ps= 0.43 x 13m/s = (      km-m/s). If we going to compare the momentum of both balls, the heavier object will surely have a greater momentum because it has a larger mass, unless otherwise  the tennis ball with a lesser mass will have a greater velocity to be equal or greater than the momentum of a soccer ball.
8 0
3 years ago
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Corrosion is the tendency of a metal to revert back to its natural state.
algol13
Corrosion is the irreversible damage or destruction of living tissue or material due to a chemical or electrochemical reaction.
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3 years ago
A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second
Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 =  \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC

Thus, the  value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0
2 years ago
Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov
Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

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Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0
2 years ago
a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.
PtichkaEL [24]

During that final period of time,
his acceleration is
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8 0
3 years ago
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