A 100-n object and a 50-n object are placed on scales a and b respectively inside an elevator ascending with constant velocity
1 answer:
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1) 5.5 N
When the ball is at the bottom of the circle, the equation of the forces is the following:
where
T is the tension in the string, which points upward
mg is the weight of the string, which points downward, with
m = 0.158 kg being the mass of the ball
g = 9.8 m/s^2 being the acceleration due to gravity
is the centripetal force, which points upward, with
v = 5.22 m/s being the speed of the ball
R = 1.1 m being the radius of the circular trajectory
Substituting numbers and re-arranging the formula, we find T:
2) 3.9 N
When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:
And by substituting the numerical values, we find
3) 2.3 N
When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is
And substituting the numerical values and re-arranging it, we find
4) 3.3 m/s
The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:
and re-arranging the equation, we find
Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.