Waves are common in nature because many different ____________ produce waves
1 answer:
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Answer:
Si logra alcanzar el bus.
Explanation:
Para poder solucionar este problema debemos de tener en cuenta que Alicia corre a velocidad constante para poder alcanzar el bus. La formula de la cinematica que tiene en cuenta la velocidad constante es la siguiente:
donde:
Xf = Ubicacion del punto donde se encuentra el bus [m]
Xo = Ubicacion desde donde esta Alicia [m]
v = velocidad constante = 5 [m/s]
t = tiempo [s]
Xf - Xo = 15 [m]
15 = 5*t
t = 3 [s]
Ahora con el tiempo podemos encontrar la velocidad del bus por medio de la siguiente ecuacion de cinematica para la aceleracion constante:
donde:
Vf = velocidad del bus despues de los 3 [s]
Vi = velocidad inicial = 0
a = aceleracion = 0.5 [m/s^2]
Vf = 0 + (0.5*3)
Vf = 1.5 [m/s]
La velocidad del bus es menor que la velocidad de Alicia, por ende Alicia alcanzara el bus.
Answer:
The magnitude of the resistive force exerted by the water is newtons.
Explanation:
By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:
(Eq. 1)
Where:
- Forward trust, measured in newtons.
- Resistive force exerted by the water, measured in newtons.
From (Eq. 1), we get that: ()
The magnitude of the resistive force exerted by the water is newtons.
Answer:
The average speed of the fielder is 5.24 m/s
Explanation:
The position vector of the ball after it was hit can be calculated using the following equation:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem.
When the ball is caught, its position vector will be (see r1 in the figure):
r1 = (r1x, 0.873 m)
Then, using the equation of the position vector written above:
r1x = x0 + v0 · t · cos α
0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²
Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:
r1x = v0 · t · cos α
0.873 m = v0 · t · sin α + 1/2 · g · t²
Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:
0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²
0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²
Solving the quadratic equation:
t = 0.03 s and t = 5.72 s.
It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.
With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:
r1x = v0 · t · cos α
r1x = 37.2 m/s · 5.72 s · cos 49.3°
r1x = 1.39 × 10² m
The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.
The average velocity is calculated as the traveled distance over time, then:
average velocity = treveled distance / elapsed time
average velocity = 30.0 m / 5.72 s = 5.24 m/s
