Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.
Explanation:
- Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
- Let
be the speed of the river's current given as 1.00 m/s.
- Note that this speed is the magnitude of the velocity which is a vector quantity.
- The direction of the swimmer is upstream.
Hence the resultant velocity is given as, 
= S — S 0
= 1.25 — 1
= 0.25 m/s.
Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.
Answer:
[1, 6, -2]
Explanation:
Given the following :
Initial Position of spaceship : [3 2 4] km
Velocity of spaceship : [-1 2 - 3] km/hr
Location of ship after two hours have passed :
Distance moved by spaceship :
Velocity × time
[-1 2 -3] × 2 = [-2 4 -6]
Location of ship after two hours :
Initial position + distance moved
[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]
= [3-2, 2+4, 4-6] = [1, 6, -2]
Answer:
The direct answer to the question as written is as follows: nothing happens to gravity when someone jumps up - gravity continues exerting a force on the body of that particular someone proportional to (mass of someone) x (mass of Earth) / (distance squared). What you might be asking, however, is what is the net force acting on the body of someone jumping up. At the moment of someone jumping up there is an upward acceleration, i.e., an upward-directed force which counteracts the gravitational force - this is the net force ( a result of the jump force minus gravity). From that moment on, only gravity acts on the body. The someone moves upward gradually decelerating to the downward gravitational acceleration until they reaches the peak of the jump (zero velocity). Then, back to Earth.
