What is a rock cycle?

Physics

1 answer:

DochEvi [55]3 years ago

7 0

Answer:

an idealized cycle of processes undergone by rocks in the earth's crust

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What are the SI units of thermal conductivity?

kotykmax [81]

Answer:

The SI unit of thermal conductivity is watts per meter-kelvin (W/(m⋅K)).

Explanation:

hope this will help u

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2 years ago

What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?

Lisa [10]

<span>Ans : Initial E = KE = Â˝mvÂ˛ = Â˝ * 1.2kg * (2.2m/s)Â˛ = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v â†’ v = 0.6 m/s which means the combined KE = Â˝ * (1.2 + 3.2)kg * (0.6m/s)Â˛ = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = Â˝kxÂ˛ = Â˝ * 554N/m * xÂ˛ x = 0.0076 m â† (a)</span>

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3 years ago

The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo

Lina20 [59]

I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J.
At point A, all this energy has converted into kinetic energy, which is:
$K= \frac{1}{2}mv^2$
And since K=7.35 J, we can find the velocity, v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s$

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3 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost

tangare [24]

The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back

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3 years ago

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
$\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}$

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3 years ago