Answer:
F = 3.86 x 10⁻⁶ N
Explanation:
First, we will find the distance between the two particles:

where,
r = distance between the particles = ?
(x₁, y₁, z₁) = (2, 5, 1)
(x₂, y₂, z₂) = (3, 2, 3)
Therefore,

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

where,
F = magnitude of force = ?
k = Coulomb's Constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 2 x 10⁻⁸ C
q₂ = magnitude of second charge = 3 x 10⁻⁷ C
r = distance between the charges = 3.741 m
Therefore,

<u>F = 3.86 x 10⁻⁶ N</u>
Answer:
Ro = 133 [kg/m³]
Explanation:
In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

where:
m = mass [kg]
V = volume [m³]
We will convert the units of length to meters and the mass to kilograms.
L = 15 [cm] = 0.15 [m]
t = 2 [mm] = 0.002 [m]
w = 10 [cm] = 0.1 [m]
Now we can find the volume.
![V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]](https://tex.z-dn.net/?f=V%20%3D%200.15%2A0.002%2A0.1%5C%5CV%20%3D%200.00003%20%5Bm%5E%7B3%7D%20%5D)
And the mass m = 4 [gramm] = 0.004 [kg]
![Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]](https://tex.z-dn.net/?f=Ro%20%3D%200.004%2F0.00003%5C%5CRo%20%3D%20133%20%5Bkg%2Fm%5E%7B3%7D%5D)
The appropriate response is letter D. The wave ventures slower and with an expanded wavelength when a sound wave entering a range of hotter air. Hotter air implies less thick, so the wave ought to back off.
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