Would you expect the bonds in ammonia to be polar covalent?
1 answer:
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Answer:
θ = 39.7º
Explanation:
In this exercise we must use Newton's second law for the sphere, at the equilibrium point we write the equations in each exercise; we will assume that plate 1 is on the left
Y Axis
-W = 0
= W
X axis
-<u> - F_{e2} + Tₓ = 0
</u>
<u> </u>
let's use trigonometry to find the components of the tension, we measure the angle with respect to the vertical
sin θ = Tₓ / T
cos θ = T_{y} / t
Tₓ = T sin θ
T_{y} = T cos θ
let's use gauss's law to find the electric field of each leaf; We define a Gaussian surface formed by a cylinder, so the component of the field perpendicular to the base of the cylinder is the one with electric flow.
F = ∫ E. dA = / ε₀
in this case the scalar product is reduced to the algebraic product, the flow is towards both sides of the plate
F = 2E A = q_{int} / ε₀
let's use the concept of surface charge density
σ = q_{int} / A
we substitute
2E A = σ A /ε₀
E = σ / 2ε₀
this is the field created by each plate. The electric force is
= q E
for plate 1 with σ₁ = -30 10⁻⁶ C / m²
F_{e1} = q σ₁ /2ε₀
for plate 2 with s2 = ab 10⁻⁶ C / m², for the calculations a value of this charge density is needed, suppose s2 = 10 10⁻⁶ C / m²
F_{e2} = q σ₂ /2ε₀
we substitute and write the system of equations
T cos θ = mg
- q σ₁ / 2ε₀ - q σ₂ /2ε₀ + T sinθ = 0
we introduce t in the second equations
- q /2 ε₀ (σ₁ + σ₂) + (mg / cos θ) sin θ = 0
mg tan θ = q /2ε₀ (σ₁ + σ₂)
θ = tan -1 (q / 2ε₀ mg (σ₁ + σ₂)
data indicates the mass of 0.25 g = 0.25 10⁻³ kg
give the charge density on plate 2, suppose ab = 10 10⁻⁶ C / m²
let's calculate
θ = tan⁻¹ (9.0 10⁻¹⁰ (30 + 10) 10⁻⁶ / (2 8.85 10⁻¹² 0.25 10⁻³ 9.8))
θ = tan⁻¹ 8.3 10⁻¹)
θ = 39.7º
Answer:
D. The more the change in gravitational potential energy.
Explanation:
From Physics, we know that change in gravitational potential energy (), measured in joules, is directly proportional to change in height (
), measured in meters. Then, the higher her jump, the more the change in gravitational potential energy.
We cannot speak of elastic energy, since statement of the problem does not speak of any deformable system (i.e. spring) in any form.
Therefore, the correct answer is D.