-- The overall <em>distance</em> he travels is (100m + 30m + 70m) = <em>200 meters</em>.
-- His <em>displacement </em>when he arrives at his front door is
D = (100m East) + (30m West) + (70m East)
D = (100m + 70m)East + (30m)West
D = (170m East) + (30m West)
<em>D = 140 meters East </em>
It's interesting to notice that his displacement is 60 meters shorter than the distance he walked.
That's because there's a stretch of 30 meters somewhere in the middle that he actually covered <em>three times</em>.
Two of those times added to the distance his shoes covered (2x30m=60m), but they cancelled out of the displacement.
His front door is 140 meters East of school. He walked 60m farther than that, going back and forth over the 30m piece.
Pretty sure it's the last one:
"reduce the size of the input force needed to perform a task"
Answer:
the Jack speed is greater than the Nas speed
v₁ = 2.08 m / s > v2 = 1.85 m / s
Explanation:
For this exercise we can find the average speed of each of the boys, the average speed is defined as the displacement in the time interval
v = x / t
Jack.
It tells us that it travels x = 15 km in a time of t = 2 h
let's reduce the magnitudes to the SI system
x = 15 km (1000 m / 1 km) = 15 10³ m
t = 2 h (3600 s / 1 h) = 7200 s
let's calculate
v₁ = 15 10³/7200
v₁ = 2.08 m / s
Nas travels a distance of x = 10 km in a time of t = 1.5 h
x = 10 km = 10 10³ m
t = 1.5 h (3600s / 1h) = 5400 s
let's calculate the speed
v2 = 10 10³/5400
v2 = 1.85 m / s
From these results we can see that the Jack speed is greater than the Nas speed
the ratio of the force produced by a machine to the force applied to it, used in assessing the performance of a machine.
Labels that belong in regions marked X and Y are;
A) X: Maintains magnetic properties in the presence of another magnet
Y: Is made from magnetically hard ferromagnetic material
