2 years ago

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

1 answer:

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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Given 1 inch ≡ 2.54 cm and 1 foot ≡

d1i1m1o1n [39]

Answer:

2991.47 [cm^2]

Explanation:

To solve this problem we must perform a dimensional analysis and use the corresponding conversion values:

$3.22[ft^{2}]*\frac{12^{2}in^{2} }{1^{2}ft^{2}} *\frac{2.54^{2}cm^{2} }{1^{2}in^{2} } \\2991.47[cm^{2}]$

3 0

3 years ago

After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample

alina1380 [7]

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

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3 years ago

At which point would most of the pendulum's potential energy have been transformed into kinetic energy?

Bess [88]

Answer:

Explanation:

i could be wrong but it seems the most logical

6 0

3 years ago

When work is done on a spring to stretch it, elastic potential energy is stored in the spring.

aniked [119]

Explanation:

I think its true

hope it helps

6 0

3 years ago

A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

$E=7*10^{9}N/C$

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field is mathematically given as

$E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\$

$E=7*10^{9}N/C$

For more information on this visit

brainly.com/question/21811998

4 0

3 years ago

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

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