We're happy that you're asking for the "displacement", because displacement is simply the straight-line distance between the start-point and end-point, and we don't care about any of the motions or gyrations along the way.
From the graph:
-- The location of the object at time-zero, when time begins, is 10 meters.
-- The location of the object after 6.0 seconds is 4 meters.
-- The distance between the start-point and end-point is
(final location) - (initial location)
-- So Displacement = (4 meters) - (10 meters)
<em>Displacement = -6 meters</em>
The answer to your question is "science"
Answer:
a) A1 = 
b) A1 = 2.688 cm
c) Q1 = A1 x v1
d) v1 = 3.1994 m/s
e) A2 = 
f) A2 = 0.7963cm
Explanation:
a) Area = 
r = 
thus,
area = 
A1 = ![\frac{\pi (d1)^{2} }{4}[/tex]b) d1 = 1.85 cmsubstituting in the above equation,A1 = [tex]\frac{\pi (d1)^{2} }{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%28d1%29%5E%7B2%7D%20%7D%7B4%7D%5B%2Ftex%3C%2Fstrong%3E%5D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3Eb%29%20d1%20%3D%201.85%20cm%3C%2Fp%3E%3Cp%3Esubstituting%20in%20the%20above%20equation%2C%3C%2Fp%3E%3Cp%3EA1%20%3D%20%20%5Btex%5D%5Cfrac%7B%5Cpi%20%28d1%29%5E%7B2%7D%20%7D%7B4%7D)
A1 = 
A1 = 2.688 cm
c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)
Q1 = A1 x v1
d) From the above equation,
v1 = 
= 
= 319.94 cm/s = 3.1994 m/s
e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.
i.e. A1 x v1 = A2 x v2
thus,
A2 =
f) v2 = 10.8 m/s.
substituting the values in the above equation,
A2 = 
= 0.7963cm
I would say that you're describing an impossible situation.
If the object is stationary, then it has no movement.
If the object has movement, then it's not stationary.
Answer:
The entire cart/hanging mass system follows the same law, ΣF = ma. This means that plotting force vs. acceleration yields a linear relationship (of the form y = mx).
