Answer:
We would take the mass given, 80 kg, and multiply it times the acceleration of Earth's downward pull on that mass, 9.8 m/s/s. So the weight on Earth is 784 newtons. So the weight of the same 80 kg mass on the moon is 133.28 N.
Answer:
a) 
b) 
c) 
Explanation:
a) In the equilibrium position of the system, that is when the spring is not elongated, the potential energy is zero. Therefore, the total energy of the system is the maximum kinetic energy:
b) The force constant of the spring can be calculated from the natural frequency of the system:
Recall that 
, that is the distance traveled in one revolution divided into the time of one revolution. Replacing and solving for k:
c) The maximum speed is directly proportional to the amplitude of the motion:
Hypothesis , I believe the answer is hypothesis
<h2>
Distance traveled in 1 second after drop is 4.9 m</h2><h2>
Distance traveled in 4 seconds after drop is 78.4 m</h2>
Explanation:
We have s = ut + 0.5at²
For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²
Substituting
s = 0 x t + 0.5 x 9.8 x t²
s = 4.9t²
We need to find distance traveled in 1 s and 4 s
Distance traveled in 1 second
s = 4.9 x 1² = 4.9 m
Distance traveled in 4 seconds
s = 4.9 x 4² = 78.4 m
Distance traveled in 1 second after drop = 4.9 m
Distance traveled in 4 seconds after drop = 78.4 m
Answer:
dx/Dt x B . x =0
Explanation:
Let's calculate the work and the magnetic force, the expression for magnetic force is
F = qv x B
Bold indicate vector quantities, the expression for the job is
W = F. X
Let's replace in this equation
W = q v x B . X
The definition of speed is
v = dX / dt
With what work is left
W = q dX / dt x B . X
As we can see the vector product gives us a vector perpendicular to dX and its scalar product by X of zero
Second part
The speed a vector and although the magnitude is constant the change of direction implies a change in the speed.
Let's calculate the magnitudes of speed (speed)
F = qv B sin θ
F = ma
q v B sin θ = ma
a = qvB / m senT
This acceleration is perpendicular to the magnetic field and the velocity, so it does not change if magnitude but its direction, it is directed to the center of the circle.
| v | = q vB/m sin θ
