Answer:
to the right.
to in the upwards direction.
Explanation:
In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).
So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:
F=ma
so in order to find the components of the force, we need to start by finding its acceleration.
Acceleration is found by using the following formula:
so we can subtract the two vectors, like this:
which yields:
or:
so now I can find the components of the force:
which yields:
F=(2.31i+2.1j)N
so the components of the force are:
to the right.
to in the upwards direction.
The position function x(t) of a particle moving along an x axis is 
a) The point at which particle stop, it's velocity = 0 m/s
So dx/dt = 0
0 = 0- 12t = -12t
So when time t= 0, velocity = 0 m/s
So the particle is starting from rest.
At t = 0 the particle is (momentarily) stop
b) When t = 0
SO at x = 4m the particle is (momentarily) stop
c) We have
At origin x = 0
Substituting
t = 0.816 seconds or t = - 0.816 seconds
So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.
Answer:
Part A. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.
Part B. The magnitude of normal force acting on the suitcase is equal to the sum of the weight of the suitcase and the man.
Explanation:
Part A. This is because when the man pulls on the suit upwards, he exerts a force in the upward direction. This takes part of the force of weight of the suitcase and decreases the force the suitcase is exerting on the ground. Thus, the normal force (force exerted by suitcase on the ground) also decreases by the same force as the pull.
Part B. The statements for this part were not given in the question, but the answer reflects what is going to happen in that scenario. Since the man sits on the suitcase, the total weight acting on the ground through the suitcase is that of the suitcase plus the man. Since this force (acting on the ground) is normal force, the statement given in the answer is correct.
To solve this problem we will use the vector concept given by the cross product between two perpendicular vectors and which results in a vector perpendicular to these two. From the definition of the Magnetic Force we have to
From the property of cross product the magnetic force should point in the direction perpendicular to the plane containing the vectors v and B.
The direction of velocity is north, and the direction of the magnetic force is northeast.
This cannot be the case, as the direction of magnetic force is not perpendicular to the direction of velocity of the charge.
Therefore the correct option for the direction of the magnetic field is <em>"This situation cannot exist because of the relative orientations of the velocity and force vectors" </em>
