Answer:
f1 = 58.3Hz, f2 = 175Hz, f3 = 291.6Hz
Explanation:
lets assume speed of sound is 350 m/s.
frequencies of a standing wave modes of an open-close tube of length L
fm = m(v/4L)
where m is 1,3,5,7......
and fm = mf1
where f1 = fundamental frequency
so therefore: f1 = 350 x 4 / 1.5
f1 = 58.3Hz
f2 = 3 x 58.3
f2 = 175Hz
f3 = 5 x 58.3
f3 = 291.6Hz
La rapidez es una magnitud escalar que relaciona la distancia recorrida con el tiempo.
La velocidad es una magnitud vectorial que relaciona el cambio de posición (o desplazamiento) con el tiempo.
A. The switch within the closed circuit is opened.
If the switch is opened current will stop flowing, as the current will meet a gap, that is not connected by a conductor.
Answer:
Explanation:
given that
Distance above the ground, s = 1.2 m
Time taken by the ball, t = 3 s
Velocity of the ball, v = 1.2/3 = 0.4 m/s
Maximum height reached by the ball is then given by the formula
H = v² / 2g
H = 0.4² / 2 * 9.8
H = 0.16 / 19.6
H = 0.0082 m or rather, 0.82 cm
Answered using calculus.
Antidifferentiated the acceleration to get velocity. Added variable c as we do not know if there was an extra number there yet.
Knowing that when time is 0, the velocity is 20, we can substitute those numbers into the equation and find that c = 20.
Now we have full velocity equation: v = 1.5t + 20
Now we substitute 4 into t to find out the velocity after 4 seconds. This gives us the final answer of 26m/s