Answer:
C
Explanation:
The weight will always be different while mass is described as the stuff inside an object, and that stays the same.
Such as it weighs differently in space.
Explanation:
a) The height of the ball h with respect to the reference line is

so its initial gravitational potential energy
is



b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

We know that the initial kinetic energy
as well as its final gravitational potential energy
are zero so we can write the conservation law as

Note that the mass gets cancelled out and then we solve for the velocity v as



Answer:
<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>
Explanation:
For a charged particle, the force on the particle is given as

also recall that work is force times the distance traveled
work = F x d
so, the work on the particle = F x d,
where the distance traveled by the particle in one revolution = 
Work on a particle = 2πrF = 
This work is proportional to the energy of the particle.
And the work is also proportional to the radius of travel of the particles.
Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.
Answer:
161.86 N
Explanation:
mass of box m= 55.0 kg
weight of the box, mg= 55×9.81
g here is acceleration due to gravity =9.81 m/sec^2
coefficient of friction between the box and the surface μ= 0.3
the friction force F_s= μmg= 0.3×55×9.81
=161.86 N
to move the ball horizontal force required is 161.86 N
The initial force between the two charges is given by:

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:

So, the new force is:

So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:

So, the new force is:

So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:

So, the new force is:

So the force has increased by a factor 6.