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asambeis [7]
1 year ago
14

The density of silver is 10.5 g/cm3g/cm3. part a what is this value in kilograms per cubic meter?

Physics
1 answer:
OLEGan [10]1 year ago
5 0

The correct answer for the density of silver in kilograms per cubic meter is 10500 kg/m^3.

Silver has a density of 10.8103 kg/m3.Pure silver is nearly white, lustrous, soft, very ductile, and malleable, and it is an excellent heat and electricity conductor. Although it is not a chemically active metal, it is attacked by nitric acid (which forms nitrate) and hot concentrated sulfuric acid.

Atomic weight: 107.87 g.mol-1

At 20°C, the density is 10.5 g.cm-3.

962 °C melting point

2212 °C is the boiling point.

Silver has long been used in the manufacture of coins, ornaments, and jewelry due to its relative scarcity, brilliant white color, malleability, ductility, and resistance to atmospheric oxidation. Silver, along with gold and the platinum-group metals, is one of the so-called precious metals.

Learn more about density of silver here :-

brainly.com/question/11025956

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Answer:

https://www.khanacademy.org/science/physics/work-and-energy/work-and-energy-tutorial/a/what-is-conservation-of-energy

Explanation:

Look at this website it gives you all the details! :)

7 0
3 years ago
you are trying to figure out the voltage of a battery when you know the current and the resistance. the current(i) is 25 Amps, a
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Answer:

V= 250 volts

Explanation:

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8 0
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Is it possible to stand backwards on a flight of stairs?
Pachacha [2.7K]

Answer:

yes

Explanation:

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7 0
3 years ago
A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt
Yuri [45]
The total power emitted by an object via radiation is:
P=A\epsilon \sigma T^4
where:
A is the surface of the object (in our problem, A=1.25 m^2
\epsilon is the emissivity of the object (in our problem, \epsilon=1)
\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4) is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is T=100^{\circ} C=373 K

Substituting these values, we find the power emitted by radiation:
P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW
So, the correct answer is D.
6 0
4 years ago
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Marat540 [252]

Answer: Its the second option

Explanation:

I have the same text and everything I had did it yesterday and picked the second one and it was right so yea. Hopefully that helps. Is this fro STEM physics

4 0
3 years ago
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