Answer:
If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.
Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:
PE = qV
where q =1.6 x 10^-19 C
and V = 1.00 V
Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:
PE* = 0,
KE* = 1/(2mv^2)
Now since
PE + KE = Total energy =PE* + KE*
Therefore,
qV + 0 = 0 + 1/2mv^2
Or
KE = qV = 1.6 10^-19 J
Answer:
When the elevator is accelerating downward, the person feels lighter due to the downward normal force being less than the person's weight.
Explanation:
A person riding in an elevator subjected to a series of unbalanced forces depending on the direction the elevator is travelling.
Two forces are acting on the person; the force of gravity and the upward normal force from the elevator.
When the elevator is going upwards with acceleration a, the person feels heavier than his normal weight, due to the upward normal force being greater than the person's weight. N = mg + ma
When the elevator is moving downwards with acceleration a, the person feels lighter due to the downward normal force being less than the person's weight. N = mg - ma
However, when the elevator is moving up or down at constant velocity ie. acceleration a = 0, the person experience a normal force equal to weight. N = mg
When the elevator is moving downwards with acceleration a = g, the person experiences weightlessness. N = (mg - mg) = 0
Answer:
D.
Explanation:
To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by
Where,
t = thickness
Wavelenght
m= is an integer
n= film/refractive index
We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>
Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm
Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile
The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile
The direction is
θ = tan-1 (-0.3/0.9)
θ = 161.57 degrees.
