Take for example driving by with a cake in your hand, then dropping it while going 30 mph. It will not drop directly down, it will gradually go in the direction you were driving while falling.
This is true I believe, if I'm interpreting correctly.
In a fluid, all the forces exerted by the individual particles combine to make up the pressure exerted by the fluid
Due to fundamental nature of fluids, a fluid cannot remain at rest under the presence of shear stress. However, fluids can exert pressure normal to any contacting surface. If a point in the fluid is thought of as a small cube, then it follows from the principles of equilibrium that the pressure on every side of this unit of fluid must be equal. but if this were not a case, the fluid would move in the directions of the resulting force, So the pressure on a fluid at rest is isotropic.
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5 seconds is a poor time to ask about, because the speed abruptly changes at exactly 5 seconds.
Up until that time, the speed has been 1 m/s. And then, at exactly 5 seconds, it becomes zero.
_________
It's also a poor question because speed is calculated from the distance covered, but the graph shows displacement, not distance. You can't really tell the distance covered from a displacement graph.
For example, if an object happens to be moving in a circle around the place where it started, then the total distance covered keeps increasing, but its displacement is constant.
Answer:
The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.
Explanation:
As from the given data
the length of the rope is given as l=30 m
the stretched length is given as l'=41m
the stretched length required is give as y=l'-l=41-30=11m
the mass is m=95 kg
the force is F=380 N
the gravitational acceleration is g=9.8 m/s2
The equation of k is given by equating the energy at the equilibrium point which is given as

Here
m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

Now the force is
or

So here F=380 N, k=630.92 N/m

So the distance is 0.602 m