A because centrifugal is to velocity to how slow or fast something is and centrifugal has expresssed as ac=v2 / r (1)<span />
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Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Responder:
20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz
Explicación:
Dado lo siguiente:
Radio (r) del círculo = 120 cm
600 revoluciones por minuto en radianes por segundo
(600 / min) * (2π rad / 1 rev) * (1min / 60seg)
(1200πrad / 60sec) = 20π rad ^ -1
Velocidad angular (w) = 20πrads ^ -1
Velocidad lineal = radio (r) * velocidad angular (w)
Velocidad lineal = (120/100) * 20πrad
Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1
C.) Período (T):
T = 2π / w = 2π / 20π = 0.1 seg
D.) Frecuencia (f):
f = 1 / T = 1 / 0.1
1 / 0,1 = 10 Hz
