Answer:
Q = 5 L/s
Explanation:
To find the flow you use the following formula (para calcular el caudal usted utiliza la siguiente formula):
![Q=\frac{V}{t}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7BV%7D%7Bt%7D)
V: Volume (volumen) = 200L
t: time (tiempo) = 40 s
you replace the values of the parameters to calculate Q (usted reemplaza los valores de los parámteros V y t para calcular el caudal):
![Q=\frac{200L}{40s}=5\frac{L}{s}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B200L%7D%7B40s%7D%3D5%5Cfrac%7BL%7D%7Bs%7D)
Hence, the flow is 5 L/s (por lo tanto, el caudal es de 5L/s)
Answer:
(a) p = 3.4 kg-m/s (b) 37.78 N.
Explanation:
Mass of a basketball, m = 0.4 kg
Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)
It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)
(a) Change in momentum = final momentum - initial momentum
p = m(v-u)
p = 0.4 (2.8-(-5.7))
p = 3.4 kg-m/s
(b) Impulse = change in momentum
Ft = 3.4
We have, t = 0.09 s
![F=\dfrac{3.4}{0.09}\\\\F=37.78\ N](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B3.4%7D%7B0.09%7D%5C%5C%5C%5CF%3D37.78%5C%20N)
Hence, this is the required solution.