Answer:
Force applied to smaller cross section is
= 82.63 N
Explanation:
As we know
where signifies the weight of the two chair in a hydraulic-lift system
And signifies the area of the two respective chairs in a hydraulic-lift system
Given -
N
Square centimeter
Square centimeter
Substituting the given values in above equation, we get -
Force applied to smaller cross section is
= 82.63 N
The given question is incomplete. The complete question is as follows.
A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.
A) If the net force acting on the particle is N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.
Enter your answers numerically separated by commas.
Explanation:
The given data is as follows.
Q =
E = 1300 N/C in the +x direction
B = 1.02 T in the +z direction
and, in the +x direction
Also,
= qE - qvB
Now, we will calculate the value of v as follows.
v =
=
v = 458.507 m/s
Using the value for velocity, we need to know which direction it's going.
You know +x direction for E, +z direction for B and +x for .
Using the right hand rule where:
your right thumb goes toward the , then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.
Thus, we can conclude that = 0, -(458.507), 0.