Answer:
<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q = 
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( = 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
The answer would be hands! hope this helps
Answer:
Option B:
A child sitting on a swing.
Explanation:
When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.
An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.
From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.
Answer:
A) The resultant force is 43.4 [N]
B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis
Explanation:
We need to make a sketch of the different forces acting on the heavy crate.
In the attached image we can see the forces and the sum of the vector with their respective angles.
Forces in the X-axis
Forces in the y-axis
![FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]](https://tex.z-dn.net/?f=FDiony%3D0%5BN%5D%5C%5CFshirley%3D%2016.5%2Asin%2830%29%3D8.25%5BN%5D%5C%5CFjoany%3D19.5%2Asin%2860%29%3D16.88%20%5BN%5D%5C%5C%5C%5CForcesy%3D0%2B8.25-16.88%3D%20-8.63%5BN%5D)
Using the Pythagorean theorem
The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.
I agree with the other comment
