1. Looking at the planet vs. eccentricity table, which two planets have the greatest eccentricity?
1 answer:
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Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.
Direction of the net force (β)
β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
Answer:
The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>
Explanation:
The directions in which the delivery truck travels are;
1) 2.8 km North = 2.8·, in vector form
2) 1.0 km East = 1.0·, in vector form
3) 1.6 km South = -1.6·, in vector form
Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;
Final displacement, Δd = 2.8· + 1.0·
+(-1.6·
) = 1.2·
+ 1.0·
Final displacement, = 1.0· + 1.2·
Where;
Δx = The displacement in the x-direction = 1.0·
Δy = The displacement in the y-direction = 1.2·
The magnitude of the resultant displacement vector is given as follows
= √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)
The magnitude of the resultant displacement vector ≈ 1.6 km
The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE
Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.