I mean, that's just a diagram... but isn't the pathway of the particle after it has left the field perpendicular to the radius of the arc?
If this is not true, I don't know how to solve this.
I realized this morning that the path of a particle after it leaves a field is actually a tangent of the arc, which means the angle is 90 degrees! I think I know how to solve this question now thanks :)
F = qvBsin(theta)
F = (2.15 x 10-6)(1.45 x 105)(0.42T)
F = 0.13N [Towards the centre]
W = F*d = Enet
(0.13N)(x) = 1/2mv2
(0.13N)(x) = 1/2(3.2 x 10-11)(1.45 x 105)2
x = 2.59m
I am 99% sure that I cannot use force since the force on the particle disappears after it leaves the field.
I reckon...
Homework Statement
http://www.solutioninn.com/physics/electricity-and-magnetism/magnetic-fields/particle-with-charge-2.15uc-and-mass-3.20-x-10-11-kg-is
I am just going to link this because I don't know how to draw a nice diagram like that and put it here. But essentially that is the question...
Ohhh... So instead of using the weight of the object, I should be using the volume of water instead..
So the equation would be buoyant force = (density of fluid)(volume of object)(g)
T.T Buoyancy is so complicated...
For the one that's resting on the bottom, other forces acting on it would be normal force exerted by the beaker then?
Oh that makes a lot of sense now.. So I should differentiate the calculations between the completely submerged object and an object in an equilibrium..
What if the object is touching the bottom of the beaker? I am aware of archimedes' principle and such but I wasn't unsure how to understand the problem when the object is touching the bottom of the beaker. Does this simply mean that the gravitational force is much greater than the bouyant force...
Homework Statement
You are given a density of a fluid, volume of the rubber stopper and density of the rubber stopper. If a rubber stopper sinks to the bottom of the beaker filled with this fluid and stays there, find the buoyant exerted on the object by the fluid.
Homework...