The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
<span>One timing problem
in using fiscal policy to counter a recession is called the “legislative lag”
it occurs between the time the time the need for fiscal action is recognized
and between the time that it is taken in action.</span>
Answer:
It's effective temperature.
Explanation:
Answer:
q₁ = -6.54 10⁻⁵ C
Explanation:
Force is a vector quantity, but since all charges are on the x-axis, we can work in one dimension, let's apply Newton's second law
F = F₁₂ + F₂₃
the electric force is given by Coulomb's law
F = k q₁q₂ / r₁₂²
let's write the expression for each force
F₂₃ = k q₂ q₃ / r₂₃²
F₂₃ = 9 10⁹ 34.4 10⁻⁶ 72.8 10⁻⁶ / 0.1²
F₂₃ = 2.25 10³ N
F₁₂ = k q₁q₂ / r₁₂²
F₁₂ = 9 10⁹ q₁ 34.4 10⁻⁶ / 0.1²
F₁₂ = q₁ 3,096 10⁷ N
we substitute in the first equation
225 = q₁ 3,096 10⁷ +2.25 10³
q₁ = (225 - 2.25 10³) / 3,096 10⁷
q₁ = -6.54 10⁻⁵ C
Answer:
In mine I said to build the boat accordingly based on what is going to be in it
Explanation:
just remember the density of the boat cannot change but the volume and the mass can
