To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,
Q = V*A
Where,
A= Cross-sectional Area
V = Velocity
The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,


Our values are given as,


Re-arrange the equation to find the first ratio of rates we have:



The second ratio of rates is



Answer:
33.33j+6.67i km/hr
Explanation:
From the law of conservation of momentum,
Applying,
mu+m'u' = V(m+m')............... Equation 1
Where m = mass of the truck, m' = mass of the car, u = initial velocity of the truck, u' = initial velocity of the car, V = Final velocity.
Note: let j represent the north, and i represent the east
From the question,
Given: m = 1500 kg, u = 60j, m' = 1200 kg, u' = 15i
Substitute these values into equation 1
1500*60j+1200*15i = V(1500+1200)
90000j+18000i = 2700V
V = (90000j+18000i)/2700
V = 33.33j+6.67i km/hr
Answer:
Found in the nucleus, Has mass of one amu
Answer:
D.The striking foot pushes backward against the ground. The friction with the ground provides an equal and opposite force forward