<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>
The current in the circuit is 5 A
Explanation:
The intensity of current is given by the equation:
where
I is the current
q is the amount of charge passing through a given point of the circuit in a time interval of t
For the cell in this problem, we have
q = 150 C is the charge
t = 30 s is the time interval
Substituting into the equation, we f ind
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Answer:
ooh thanks but don't give me advise
Answer:
b) directional
Explanation:
The prediction that alcohol slows reaction time is Directional . This is because for alcohol to be known as one which slows reaction time means there have been various hypothesis conducted which supports this.
In this case, formulation of a null hypothesis is usually necessary which means that alcohol does not slow reaction time and another alternative hypothesis that suggests that alcohol slows reaction time.
Answer:
F=5833.3 N N
Explanation:
Newton's second law applied to the car
F= m*a Formula (1)
F: Force in Newtons (N)
m : mass in kg
a: acceleration ( m/s²)
kinematics car
vf= v₀ + a*t Formula (2)
vf : final velocity (m/s)
v₀ : final velocity (m/s)
a : acceleration ( m/s²)
t : time t
Equivalences
1 km= 1000m
1 h = 3600 s
Data
m= 1000kg
v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s
vf= 0
t= 6 s
Problem Development
We calculate the acceleration replacing the data in the formula (2) :
0 = 25 + a*6
a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)
We calculate the force is required to stop the car replacing the data in the formula (1)
-F = 1400 kg*(-4.16 m/s²)
F=5833.3 N
