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beks73 [17]
3 years ago
13

PLZ HELP ME!!!!!!!!! WILL GIVE BRAINLY!!!!!!!! Which photo represents a waxing gibbous moon?

Physics
1 answer:
yarga [219]3 years ago
6 0
A phase of moon with more than half sunlit portion visible on the right
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3 years ago
Two cars are raised to the same elevation on service- station lifts. If one car is twice as massive as the other, how do their p
Brums [2.3K]

Answer:

The potential energy of the more massive one is twice that of the other.

Explanation:

Potential energy is given by

<em>PE</em> = <em>mgh</em>

where <em>m</em> = mass of body, <em>g</em> = acceleration of gravity and <em>h</em> = height or elevation.

For the less massive car, let the mass be m_1. Then its <em>PE</em> is

PE_1 = m_1gh

For the massive car, let the mass be m_2.  Its <em>PE</em> is

PE_2 = m_2gh

But m_2 =2m_1

\therefore PE_2 = 2m_1gh = 2(m_1gh) = 2PE_1

Hence, the potential energy of the more massive one is twice that of the other.

7 0
3 years ago
Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su
dybincka [34]
  1. La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
  2. La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua (v), en metros por segundo:

v = \sqrt{\frac{K}{\rho} } (1)

Donde:

  • K - Módulo de compresibilidad, en newtons por metro cuadrado.
  • \rho - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que \rho = 1\times 10^{3}\,\frac{kg}{m^{3}} y K = 2,16\times 10^{9}\,\frac{N}{m^{2}}, entonces la velocidad de las ondas sonoras es:

v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }

v\approx 1469,694\,\frac{m}{s}

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda (\lambda), en metros, mediante la siguiente fórmula:

\lambda = \frac{v}{f} (2)

Donde f es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que v\approx 1469,694\,\frac{m}{s} y f = 1000\,hz, entonces la longitud de onda de las ondas sonoras es:

\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}

\lambda = 1,470\,m

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

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A cd player uses a ___ to convert info to an electrical signal
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C.) Laser. the light from the laser reflects off the shiny surface as the CD rotates
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attashe74 [19]

Answer: 161.3

I have a acellus too and got this question correct, so I hope this helps y’all out

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