Answer:
F = 5.83 10⁻¹⁷ N
Explanation:
The electric force is given by
F = k q₁ q₂ / r²
With Gauss's law electric field flow is equal to the charge inside the Gaussian surface, if we make a spherical surface around each drop, the force independent of small deformations due to air resistance
q₁ = q₂
F = 8.99 10⁹ (29 10⁻¹²)² / (0.36 10⁻²)²
F = 5.83 10⁻¹⁷ N
As the two drops have a charge of the same sign they repel
Answer:
Explanation:
We use the harmonic motion position equation:
where A = 0.350 and for t = 0
so:
and also:
so we have:
x(t)=0.350cos(1.532 t)
For t = 3.403 s
x(3.403)=0.350cos(1.532 (3.403)) = 0.348 m
Jupiter, Saturn, Uranus and Neptune are the jovian planets.
Mercury, Venus and Earth<span> are the terrestrial planets.
</span>One of the main differences that can be seen between terrestrial and jovian planets, is their surfaces. While the terrestrial planets are made of solid surfaces, the jovian planets are made of gaseous surfaces.
The jovian planets are less dense compared to the terrestrial planets, because they are mainly composed of hydrogen gas. Moreover, the core of the jovian planets is more dense than the terrestrial planets.<span>
</span>So, <span>Terrestrial planets are mainly composed of _____solid surfaces_______ while jovian planets are made dominantly of __gaseous______.</span>
A compound contains two or more different elements.
It also contains two or more atoms.
Both answers work
Answer:
x = 41.2 m
Explanation:
The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.
In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract
F = F₁₃ - F₂₃
let's write the expression for each force, let's set a reference frame on the charge q1
F₁₃ =
F₂₃ =
they ask us that the net force be zero
F = 0
0 = F₁₃ - F₂₃
F₁₃ = F₂₃
k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}
q1 / x2 = q2 / (d-x) 2
(d-x)² = x²
we substitute
(100 - x)² = 2/1 x²
100- x = √2 x
100 = 2.41 x
x = 41.2 m