QUESTION:
WHAT IS THE MAGNITUDE OF THE MAGNETIC FIELD AT RIGHT ANGLES TO THE PROTON'S PATH?
ANSWER:
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Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
The simplest answer would be "acceleration due to gravity."
The exact value of this acceleration changes depending on which planet your on (for example).
Answer:
y = 33.93 10⁵ m
Explanation:
This is an interference exercise, for the contributory interference is described by the expression
d sin θ = m λ
let's use trigonometry for the angle
tan θ = y / L
how the angles are small
tan θ = sin θ / cos tea = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
the light fulfills the relation of the waves
c = λ f
λ = c / f
λ = 3 10⁸ /375
λ = 8 10⁵ m
first order m = 1
let's calculate
y = 1 8 10⁵ 4030 10-9 / 950 10-9
y = 33.93 10⁵ m