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Alex_Xolod [135]
3 years ago
5

In the equation for elastic potential energy below, what is represented by the symbol k? Ee = ½ × k × e²

Physics
1 answer:
yarga [219]3 years ago
5 0

Answer:

K is the spring/elastic constant measured in Newton per metre square (N/m²)

Explanation:

E = ½Ke²

In the equation above,

E is the energy in Joules (J)

K is the elastic/spring constant in Newton per metre square (N/m²)

e is the extension in metre (m)

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A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra
son4ous [18]

Answer:

The answer is v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}.

Explanation:

From the law of gravity,

F = G \frac{Mm}{r^2}

considering F as a conservative force, F = - \nabla U,

the general expression for gravitational potential energy is

U = -G \frac{Mm}{r},

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

g = -G \frac{M}{r^2},

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

U = g M_e R,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

U_p = g M_e(R- r_s),

then, the kinetik energy when the rock hits the surface is

U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s),

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g = G \frac{M_s}{R^2}.

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Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air
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The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

f=\frac{v}{\lambda}

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Applying the formula:

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f=\frac{400 m/s}{2 m}=200 Hz

- underwater, the frequency of the wave is:

f=\frac{1600 m/s}{8 m}=200 Hz

So, the frequency has not changed.

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3 years ago
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denpristay [2]

Answer:

Person B

Explanation:

Person B will hear the frequency that is lower than the B- flat.

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