The original Coulomb force between the charges is:
Fc=(k*Q₁*Q₂)/r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.
The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive.
Q₁ is decreased to Q₁₁=(1/3)*Q₁=Q₁/3 and
Q₂ is decreased to Q₂₂=(1/2)*Q₂=Q₂/2.
New force:
Fc₁=(k*Q₁₁*Q₂₂)r², now we input the decreased values of the charge
Fc₁=(k*{Q₁/3}*{Q₂/2})/r², that is equal to:
Fc₁=(k*(1/3)*(1/2)*Q₁*Q₂)/r²,
Fc₁=(k*(1/6)*Q₁*Q₂)/r²
Fc₁=(1/6)*(k*Q₁*Q₂)/r², and since the original force is: Fc=(k*Q₁*Q₂)/r² we get:
Fc₁=(1/6)*Fc
So the magnitude of the new force Fc₁ with decreased charges is 6 times smaller than the original force Fc.
Answer:
<em>The pressure given by the water at the bottom is 58,800 Pa</em>
Explanation:
<u>Pressure Exerted by a Column of Liquid
</u>
The pressure P exerted by a column of liquid of height h and density ρ is given by the hydrostatic pressure equation:
P = ρgh
Where ρ is the density of the liquid, g is the acceleration of gravity g=9.8m/s^2, and h is the height of the column of liquid.
The height of water in a well is h=6 m. The density of water is a known quantity:
Calculating P:
P = 1,000*9.8*6
P = 58,800 Pa
The pressure given by the water at the bottom is 58,800 Pa
1. The wavelength is the ratio of the wave's speed to its frequency in hertz or 1/s. This is shown below,
λ = s / f = (320 m/s) / (300 1/s) = 1.07 m
The wavelength is approximately 1.07 m.
Nucleus? that's what I'm guessing but I'm not sure it's right :/
Explanation: Formulas you need to use:
K=0.5mass(Velocity^2)
km/h / 3.6 ===>M/S
Step 1:
convert km/h to m/s
Put the values in the formula :