Answer:
D. move up to another shell that would form
Explanation:
An atom has protons, neutrons and electrons. Protons and neutrons are present in the nucleus and electrons orbit the nucleus in fixed shells. An electron can jump to higher shell when it gains energy and lower one when it loses energy. Thus, when single electron in hydrogen atom is given a small amount of energy, it would jump to another higher shell.
Answer:
Impedance = 93.75 ohms
Current = 1.81 A
Explanation:
Resistance = R = 80 ohms
Inductance = L = 0.2 H
Inductive reactance = XL = 
= ωL = (2πf) L
= 2 (3.14) (60)(0.2) = 75.398 Ohms
Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]
= 26.526 Ohms
Impedance = Z = 
= 
= 93.747 ohms
Voltage = 
× 120 = 169.7056 V
Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A
Answer:
P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,
Explanation:
This problem of fluid mechanics let's start with the continuity equation to find the speed of water output
Q = A v
v = Q / A
The area of a circle is
A = π r² = π d² / 4
Let's look at the speeds at each point
v₁ = Q / A₁ = Q 4 /π d₁²
v₁ = 10 4 /π 0.5²
v₁ = 50.93 m / s
v₂ = Q / A₂
v₂ = 10 4 /π 0.25²
v₂ = 203.72 m / s
Now we can use Bernoulli's equation in the colon
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear
P₁ = P2 + ½ rho (v₂² - v₁²)
P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)
P₁ = 1.013 10⁵ + 2.205 10⁷
P₁ = 2.215 10⁷ Pa
la definicion de presion es
P₁ = F₁/A₁
F₁ = P₁ A₁
F₁ = 2.215 10⁷ pi d₁²/4
F₁ = 2.215 10⁷ pi 0.5²/4
F₁ = 4.3 106 N
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
