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finlep [7]
2 years ago
9

How do density, pressure, and temperature change as the depth of Earth increases? Question 5 options: Density, pressure, and tem

perature all increase as you go deeper into the Earth. Density, pressure, and temperature all decrease as you go deeper into the Earth. Density decreases, and pressure and temperature increase as you go deeper into the Earth. Density and pressure decrease, and temperature decreases as you go deeper into the Earth.
Physics
1 answer:
Daniel [21]2 years ago
3 0

Answer: Density, pressure, and temperature all increase as you go deeper into the Earth.

Explanation: Explanation: As you go deeper in depth, pressure increases. Density = mass/volume. The layers beneath us due to pressure get packed to the point of being very dense.

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A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
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Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

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Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

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The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

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So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

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Putting the given values into the above formula as follows.

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Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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