Answer:
273.15 dat is the answer im pretty sure
<span>R = rate of flow = 0.370 L/s
H = height = 2.9 m
T= time = 3.9 s
V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2
G value = 9.8 m/s2
Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N
Wa = weight of accumulated water after 3.9 s
Fi = force of impact of water on the bucket
S = reading on the scale = Wa + Wb + Fi
mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg
Therefore, Wa = 1.443 x 9.8 = 14.1414 N
Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt)
= 0.37 x 7.539 = 2.78943 N
S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
Answer:
F= 600 N
Explanation:
Given that
Initial velocity ,u= 0 m/s
Final velocity ,v= 30 m/s
mass ,m = 0.5 kg
time ,t= 0.025 s
The change in the linear momentum is given as
ΔP= m (v - u)
ΔP= 0.5 ( 30 - 0 ) kg.m/s
ΔP= 15 kg.m/s
We know that from second law of Newtons
Now by putting the values
F= 600 N
