Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of
1 answer:
<u>Answer</u>
The combined displacement is 2km north
<u>Explanation</u>
Since displacement is a vector quantity, we take into account the direction.
Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.
We now add to obtain,
This will simplify to
Therefore the combined displacement is 2km north
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Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity will be expressed by the following equation:
Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.
Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s