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natali 33 [55]
2 years ago
12

How to find density of a gas.

Physics
1 answer:
antoniya [11.8K]2 years ago
3 0

Answer:

Subtract the mass of the cylinder from the mass of the cylinder when it contains gasoline. This is the mass of the gasoline. Divide this figure by the volume, 100 ml, to get the density.

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Que significa gobernar​
kow [346]

Answer: conducir la política, acciones y asuntos de (un estado, organización o personas).

7 0
3 years ago
Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109
Oksana_A [137]

Answer:

 A) F = 1.09 10 5 N, b) Yes  

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

               ρ = m / V

               .m = ρ V = ρ 4/3 π r³

The radius is half the diameter

               r = 1.9 10⁻² / 2 = 0.95 10⁻² m

               ρ = 8960 kg / m3

               m = 8960 4/3 π (0.95 10⁻²)³

               m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

             #_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

             #_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

               q = e / 10⁹    #_atom

               q = e / 10⁹    3,049 1023

               q = 3,049 10⁴  (-1.6 10⁻¹⁹)

               q = -4,878 10-5 C

Electric force is

             F = k q₁q₂ / r²

             F = k q² / r²

             

Let's calculate

             F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

              F = 1.09 10 5 N

This is a force of repulsion.

Part B

 The magnitude of this force is  in very easy to detect

4 0
3 years ago
Suppose we have a laser emitting a diffraction-limited beam (=632 nm) with a 2-mm diameter. How big a light spot from this lase
expeople1 [14]

Answer:

Explanation:

λ = wave length = 632 x 10⁻⁹

slit width a = 2 x 10⁻³ m

angular separation of central maxima

= 2 x λ /a

= 2 x 632 x 10⁻⁹ / 2 x 10⁻³

= 632 x 10⁻⁶ rad

width in m of light spot.

= 632 x 10⁻⁶  x 376000 km

= 237.632 km

5 0
3 years ago
How are the amplitude and energy of a wave related
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I would go with the last two because they are the same  please give me brainiest if i am right

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3 0
3 years ago
Read 2 more answers
Which description does the phrase "mechanical advantage" most closely relate to?
yKpoI14uk [10]
<span>the ratio of the force produced by a machine to the force applied to it, used in assessing the performance of a machine. I would say the answer is D, but i'm not sure. :)</span>
4 0
3 years ago
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