<h3>Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva?</h3>
- Es normal que los quemadores superiores de una estufa eléctrica o una estufa se enciendan y apaguen en configuraciones distintas a Hi. El quemador se encenderá y apagará más de lo normal cuando se utilizan cacerolas que no son planas o que son del tamaño incorrecto para el quemador.
<h2>☆彡Hanna</h2>
#CarryOnLearning
348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.
To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations and where is distance, is the initial velocity, is the final velocity, is time, and is aceleration.
Superman's initial velocity is , and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know , and , we have to find the aceleration in order to find .
From the equation we have to clear , getting the equation as follows: .
Substituting the values:
To find we use the equation .
Substituting the values:
Answer:
its good no need to change anything :))
Answer:
Q = 47.06 degrees
Explanation:
Given:
- The transmitted intensity I = 0.464 I_o
- Incident Intensity I = I_o
Find:
What angle should the principle axis make with respect to the incident polarization
Solution:
- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:
I = I_o * cos^2 (Q)
- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:
Q = cos ^-1 (sqrt (I / I_o))
- Plug the values in:
Q = cos^-1 ( sqrt (0.464))
Q = cos^-1 (0.6811754546)
Q = 47.06 degrees
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal