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AURORKA [14]
3 years ago
15

the same amount of heat energy is added to equal masses of lead, iron, basalt, and water at room temperature. Assuming no phase

change takes place, which substance will have the smallest change in temperature?
Physics
1 answer:
Bumek [7]3 years ago
7 0
After equal masses absorb equal amounts of heat, the substance
with the greatest 'specific heat capacity' will have the smallest change
in temperature.

The specific heat capacities of those substances are ...

Water . . . . .  4,181       joules per kilogram-°C
Lead . . . . . . . . 125.6
Iron . . . . . . . . . 460.5
Basalt . . . . . . . . . 0.84

It looks like water is the easy winner.

THAT's why, in the days before electric blankets, hot-water bottles
were used to warm up a cold bed ... not hot-iron bottles or hot-basalt
bottles.  A pound of hot water brings much more heat to the sheets 
than a pound of any of those others.
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Sunny_sXe [5.5K]

Answer:

<h2>289.9 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

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From the question we have

momentum = 130 × 22.3

We have the final answer as

<h3>289.9 kg.m/s</h3>

Hope this helps you

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2 years ago
Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

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