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3 years ago

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When a rock is held above the ground, we say it has some potential energy. When we let it go, it falls and we say the potential

energy is converted to kinetic energy. Finally, the rock hits the ground (and stays there). What has happened to the energy?

1 answer:

snow_lady [41]3 years ago

8 0

Answer:

Energy converted to heat and sound. Deformation of soil or ground may also take up some energy.

Explanation:

When the rock hits the ground, a part of the energy is converted to heat (thermal energy) and sound. If the ground is soft, it will be compressed by the rock when it lands. This deformation of the ground will take up some of the energy.

We can think of kinetic energy being transferred to the ground in the following way:

1. Ball hits the ground, and transfers kinetic energy to lots of soil particles.

2. Soil particles move a bit and transfer the energy to soil particles ahead of them.

3. Kinetic energy is transferred to a very large number of soil particles, and they move so little that we consider this vibration motion.

4. This small vibration motion is now considered as internal energy or 'heat'.

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The gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the obj

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<h2>Weight of astronaut 2450 miles above the Earth is 80.38 pounds</h2>

Explanation:

Given that gravitational force, F, between an object and the Earth is inversely proportional to the square of the distance from the object and the center of the Earth.

$F=\frac{k}{r^2}$

Where F is gravitational force between an object and the Earth, r is the distance from the object and the center of the Earth and k is a constant.

Radius of Earth = 4000 miles

In case 1 an astronaut weighs 209 pounds on the surface of the Earth,

$209=\frac{k}{4000^2}\\\\k=3.344\times 10^9$

Now we need to find weight of astronaut 2450 miles above the Earth

r = 4000 + 2450 = 6450 miles

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Weight of astronaut 2450 miles above the Earth is 80.38 pounds

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3 years ago

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate

SIZIF [17.4K]

Answer:

$\epsilon_2=0.098$

Explanation:

Diameter $d=30cm=0.3m$

Temperature $T=600k$

Rate of supply $r=65W$

Emissivity of base surface $\in_b =0.55$

Temperature at base $T_b=400k$

Generally the equation for Area of base surface is mathematically given by

$A_b=\frac{\pi}{4}d^2$

$A_b=\frac{\pi}{4}0.3^2$

$A_b=0.0707m^2$

Generally the equation for Area of Hemispherical dome is mathematically given by

$A_h=\frac{\pi}{2}d^2$

$A_h=\frac{\pi}{2}0.3^2$

$A_h=0.1414m^2$

Since base is a flat surface

$F_{11}+F_{12}=1$

$F_{11}=0$

Therefore

$F_{12}=1$

$A_b=0.0707m^2$

Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

$Q_{21}=-Q_{12}$

$Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }$

Where

$\sigma=5.67*10^{-8}$

Therefore

$65=\frac{(5.67*10^{-8}(400^4-600^4))}{\frac{1-0.55}{0.0707*0.55}+\frac{1}{0.0707}+\frac{1-\epsilon_2}{0.1414*\epsilon_2}}$

$\epsilon_2=0.098$

$\epsilon_2 \approx 0.1$

Therefore the emissivity of the dome is

$\epsilon_2=0.098$

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3 years ago