Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system

where m is mass of the body attached with spring
Putting the values

ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
Time taken by proton to complete one complete circular orbit= 7.28 x 10⁻⁸ s
Explanation:
For proton, the centripetal force required for circular motion is provided by the magnetic force,
so Fm= Fc
q v B = m v²/r
m= mass of charged particle
v= velocity
B =magnetic field
q= charge
r= radius of circular path
v= q B r/m
now v= r ω
ω= angular velocity
ω r = q B r /m
ω=q B /m
now ω= 2π/T where T =time period
so 2π/T=q B/m
T= 2 πm/q B
T= 2π (1.67 x 10⁻²⁷)/ [( 1.6 x 10⁻¹⁹)* (0.9)]
T= 7.28 x 10⁻⁸ s
The electric field generated by a point charge is given by:

where

is the Coulomb's constant
Q is the charge
r is the distance from the charge
We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them.
Let's calculate first the electric field generated by the positive charge at that point:

where the positive sign means its direction is away from the charge.
while the electric field generated by the negative charge is:

where the negative sign means its direction is toward the charge.
If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
Answer:
Tension of 132N
Explanation:
We need to apply Summatory of Force to find the tension in the hand.
We define te tensión in the hand as
and the Tension in fence post as
, then


We apply summatory of moments then

Where the Force 2 is 1.25m from the center of summatory,
We can note that,

We have two equation and two incognites, then replacing (1) in (2)




Here, height is given which will be the distance for a freely falling object.
The velocity will be

and the acceleration will be

In this way, the formula works.