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luda_lava [24]
3 years ago
14

Accurate forecasts require careful study of the location, size, movement, and characteristics of large _____.

Physics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer: B air masses

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Assuming that each nucleus is roughlyspherical and that its mass is roughly equal to A (in atomic mass units {\rm u}), what is t
lara [203]

Answer:

ρ/ρ2 = 3 / R₀       the two densities are different

Explanation:

Density is defined as

       ρ = M / V

As the nucleus is spherical

       V = 4/3 π r³

Let's replace

      ρ = A / (4/3 π R₀³)

      ρ = ¾ A / π R₀³

b)

      ρ2 = F / area

The area of ​​a sphere is

     A = 4π R₀²

     ρ2 = F / 4π R₀²

     ρ2 = F / 4π R₀²

Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.

Let's look for the relationship of the two densities

     ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)

     ρ /ρ2 = 3 (A / F) (1 / R₀)

In this case it does not say that the nucleon number is A (F = A), the relationship is

     ρ/ρ2 = 3 / R₀

I see that the two densities are different

3 0
3 years ago
If the mass of an object is 10 Kg and it experiences a Net force of 60 Newtons. What is the acceleration?
Solnce55 [7]

Answer:

6 m/s^2

Explanation:

F = ma

60N = (10 kg)a

a = 6m/s^2

3 0
2 years ago
As the energy of a wave increases its ________ increase
weqwewe [10]
First one is B, second one is A
6 0
3 years ago
What is matter and explain​
Mice21 [21]
Matter is any substance that has mass and takes up space by having volume.
3 0
3 years ago
Read 2 more answers
The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
ivolga24 [154]

Answer: 0.10233nm

Explanation:

The mean free path \lambda   of an atom is given by the following formula:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

R=8.3145J/mol.K is the Universal gas constant

T=0\°C=273.115K is the absolute standard temperature

d is the diameter of the helium atom

N_{A}=6.0221(10)^{23}/mol is the Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} absolute standard pressure

Knowing this, let's find d from (1), in order to find the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius is half the diameter:

r=\frac{d}{2}  (5)

Then:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

However, we were asked to find this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Finally:

r=0.10233nm This is the radius of the helium atom in nanometers.

5 0
3 years ago
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