is to the east. (i) Automobile A travels 50 km due east. (ii) Automobile B travels 50 km due west. (iii) Automobile C travels 60
1 answer:
Complete Question:
Each of the following trips lasts exactly one hour. The positive direction is to the east.
(i) Automobile A travels 50 km due east. (ii) Automobile B travels 50 km due west. (iii) Automobile C travels 60 km due east, then turns around and travels 10 km due west. (iv) Automobile D travels 70 km due east. (v) Automobile E travels 20 km due west, then turns around and travels 20 km due east. (a) Rank the five trips in order of average x-velocity from most positive to most negative. (b) Which trips, if any, have the same average x-velocity? (c) For which trip, if any, is the average x-velocity equal to zero?
Answer:
a) 1. (iv) 2.) (i) and (iii). 3. (v) 4.(ii)
Explanation:
By definition, the average velocity, can be expressed as the total displacement (the difference between the final and initial position), divided by the time interval during which we define the displacement.
We can put these words in an equation format as follows:
If we choose x₀ = 0 and t₀ = 0, the average velocity is just the final position over one hour.
The final positions, in the five trips, are as follows:
(i) = +50 Km (2nd)
(ii) = -50 Km (4th)
(iii) + 60 Km -10 Km = + 50 km (2nd)
(iv) = + 70 km (1st)
(v) = -20 km + 20 Km = 0 (3rd)
b) As stated above, the trips (i) and (iii) have the same average x-velocity.
c) As stated above, in trip (v) as the total displacement is 0, average -velocity is 0 also.
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Answer:
A) 49,050 N
B) 16 m
Explanation:
Question:
El dibujo de la pregunta se obtiene de un documento titulado "TRABAJO DIVERSO Y ENERGÍA" que se encuentra en línea y se presenta aquí.
La masa dada del vagón, m = 1,000 kg
La altura del punto en el que descansa el vagón, punto 1, h₁ = 12 m
A) El radio en el punto 2, el punto más bajo, R = 6 m
La fuerza, 'N', que la vía ejerce sobre el vagón en el punto 1 viene dada por la siguiente relación;
N = El peso del vagón + La fuerza de movimiento del vagón
∴ N = m × g + m × a
Dónde;
g = La aceleración debida a la gravedad ≈ 9,81 m / s²
a = La aceleración del vagón
Observamos que para el movimiento circular, la fuerza de movimiento del vagón, m × a = La fuerza centrípeta que actúa sobre el vagón = m × v² / R
∴ m × a = m × v² / R
Dónde;
v² = La velocidad del vagón en el punto 2 = 2 · g · h₁
Por lo tanto;
N = m × g + m × a = m × g + m × v² / R = m × g + m × 2 · g · h₁ / R
∴ N = 1000 × 9,81 + 1000 × 2 × 9,81 × 12/6 = 49,050
La fuerza que ejerce el vagón en el punto 2, N = 49,050 N
B) En el punto 3, tenemos;
N = m · g - m · a₃
La fuerza centrípeta en el punto 3, m · a₃ = m · v₃² / R₃
∴ La altura en el punto 3, h₃ = 4 m
El cuadrado de la velocidad en el punto 3, v₃² = 2 · g · (h₁ - h₃)
Para que el vagón esté seguro en el punto 3, la fuerza de la vía sobre el vagón, N = 0 para que el vagón permanezca en la vía actuando
Por lo tanto;
N = m · g - m · a₃ = 0
m · g = m · a₃ = m · v₃² / R₃ = m · (2 · g · (h₁ - h₃)) / R₃
∴ R₃ = (2 · g · (h₁ - h₃)) / g = (2 · (h₁ - h₃)) = 2 × (12 - 4) = 16
El radio de curvatura en el punto 3 para que el punto sea seguro es R₃ = 16 m.
Answer:
24.21 m/s
Explanation:
x = Displacement in x direction = 55 m
y = Displacement in x direction = 0
= Height of the ball when the ball is kicked = 1 m
t = Time taken = 4.2 s
= Acceleration due to gravity =
u = Initial velocity of ball
Displacement in x direction is given by
Displacement in y direction is given by
The ball should be kicked at an angle of
The initial speed of the ball is 24.21 m/s.