Answer:
15.106 N
Explanation:
From the given information,
The weight of the bucket can be calculated as:
[tex]W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N[/tex]
The mass of the water accumulated in the bucket after 3.20s is:
[tex]m_w= (0.20 \ L/s) ( 3.20)s[/tex]
[tex]m _w=0.64 \ kg[/tex]
To determine the weight of the water accumulated in the bucket, we have:
[tex]W_w = m_w g[/tex]
[tex]W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)[/tex]
[tex]W_w = 6.272 \ N[/tex]
For the speed of the water before hitting the bucket; we have:
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}[/tex]
v = 8.4 m/s
Now, the force required to stop the water later when it already hit the bucket is:
[tex]F = v ( \dfrac {dm}{dt} )[/tex]
[tex]F = (8.4 \ m/s)( 0.200 \ L/s)[/tex]
F = 1.68 N
Finally, the reading scale is:
[tex]F_{scale[/tex] = 7.154 N + 6.272 N + 1.68 N
= 15.106 N
Answer:
F_scale ≈ 15.12 N
Explanation:
We are given;
Mass flow rate; m' = 0.2 l/s
Time; t = 3.2 s
Mass of bucket; m_b = 0.730 kg
Height; h = 3.6 m
Now, mass of water is;
m_w = 0.2 l/s × 3.2 s
m_w = 0.64 l
From conversion, 1 litre = 1 kg
Thus: m_w = 0.64 kg
Now, let's calculate final velocity from Newton's third equation of motion.
v² = u² + 2gh
Initial velocity is 0. Thus;
v² = 0 + 2(9.8 × 3.6)
v² = 70.56
v = √70.56
v = 8.4 m/s
Now, total mass of water and bucket is;
m_t = m_w + m_b = 0.64 + 0.73
m_t = 1.37 kg
Force on the scale is calculated from;
F_scale = (m_t)g + (m_w)v/t
F_scale = (1.37 × 9.81) + (0.64 × 8.4/3.2)
F_scale ≈ 15.12 N
