Answer:Fob= 86.2 lb, Foc= 175 lb, Fod= 506 lb
Explanation:firstly we need to get the spring stretch by subtracting the unstretched length from 1. The result will be (1-0.2=0.8ft). We then get the spring force by multiplying kOA by 0.8ft. therefore the spring force will be {350x0.8}= 280 lb. After that we can resolve each force as shown in the attachment.
Answer:
Value of electric field along the axis and equitorial axis and
respectively.
Explanation:
Given :
Distance between charges ,
Magnitude of charges ,
Dipole moment ,
Case A) (x,y) = (12.0 cm, 0 cm) :
Electric field of dipole in its axis ,
Putting all values and
We get ,
Case B) (x,y) = (0 cm, 12.0 cm) :
Electric field of dipole on equitorial axis ,
Putting all values and
We get ,
Hence , this is the required solution.
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
Substituting,
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
Where here is the velocity after the collision.