Bruce pulls a spring with a spring constant k=100 Nmk=100\, \dfrac{\text N}{\text m}k=100mNk, equals, 100, start fraction, star
1 answer:
Answer:
K_{e} = 2.0 J
Explanation:
In this exercise you are asked to calculate the elastic potential energy of a spring
= ½ k x²
where k is the spring constant and x is the displacement from equilibrium position
In this exercise, indicate that the spring constant is k = 100 N/m, the length at rest is x₀ = 20 cm = 0.20 m, up to the position x₁ = 40 cm = 0.40 m, therefore the elongation
Δx = x₁ - x₀
Δx = 0.40 - 0.20
Δx = 0.20 m
let's calculate the elastic potential energy
K_{e} = ½ 100 0.20²
K_{e} = 2.0 J
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Hi there!
We can use Biot-Savart's Law for a moving particle:
B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)
q = charge of particle (1.6 × 10⁻¹⁹ C)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
r = distance from particle (2.10 μm)
There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:
Where 'θ' is the angle between the velocity and radius vectors.
a)
To find the angle between the velocity and radius vector, we find the complementary angle:
θ = 90° - 60° = 30°
Plugging 'θ' into the equation along with our other values:
b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.
c)
In this instance, the radius vector and the velocity vector are perpendicular so
'θ' = 90°.
d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.
Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.