To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.
For the given problem we have to
In this way the torque acting on the particle as a function of distance and time is,
The net torque acting on the particle is
PART B) The direction of the torque is given by,
Therefore the torque direction is 48.04° below the x axis.
Answer:
341.46miles
Explanation:
Find the diagram attachment.
To get the displacement D, we will use the cosine rule as shown;
D² = 200²+150²-2(160)(400)cos65°
D² = 40000+22500-128000cos65°
D² = 62500+54095.14
D² = 116595.14
D = √116595.14
D= 341.46 miles
Hence the plane final displacement is 341.46miles
The x -component of the object's acceleration is 2 m/s².
<h3>What's the resultant force along x- direction?</h3>Thus, we can conclude that the acceleration along x axis is 2 m/s².
Disclaimer: The question was given incomplete on the portal. Here is the complete question.
Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?
Learn more about the acceleration here:
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