What determines whether an object will scatter light or reflect light?
1 answer:
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Answer:
a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s
Explanation:
a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration
Fe + Fm = m a
a = (Fe + Fm) / m
the electric force is
Fe = q E k ^
Fe = 4.25 10-4 60 k ^
Fe = 2.55 10-2 k ^
the magnetic force is
Fm = q v x B
Fm = 4.25 10⁻⁴
fm = 4.25 10⁻⁴ (-j ^ 30 4)
fm 0 = ^ -5,10 10⁻² j
We look for every component of acceleration
X axis
aₓ = 0
there is no force
Axis y
ay = -5.10 10²/5 10⁻⁵ j ^
ay = -1.02 107 j ^ / s2
z axis
az = 2.55 10⁻² / 5 10⁻⁵ k ^
az = 5.1 10² k ^ m / s²
Having the acceleration in each axis we can encocoar the position using kinematics
X axis
the initial velocity is vo = 30 m / s and an initial position xo = 0
x = vo t + ½ aₓ t₂2
x = 30 0.02 + 0
x = 0.6m
Axis y
acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m
y = I + go t + ½ ay t²
y = 1 + 0 + ½ (-1.02 10⁷) 0.02²
y = 1 - 2.04 10³
y = -2039 m j ^
z axis
acceleration is aza = 5.1 10² m / s², the position and initial speed are zero
z = zo + v₀ t + ½ az t²
z = 0 + 0 + ½ 5.1 10² 0.02²
z = 1.02 10⁻¹ m k ^
therefore the position of the bodies is
r = (0.6 i- 2039 j ^ + 0.102 k⁾ m
b) x axis
since there is no acceleration the speed remains constant
vₓ = 30.0 m / s
Axis y
let's use the equation v = v₀ + t
= 0 + -1.02 10⁷ 0.02
v_{y} = 2.04 10⁵ m / s
z axis
v_{z} = vo + az t
v_{z} = 0 + 5.1 10² 0.02
v_{z} = 1.02 10⁻¹m / s
As per Newton's 2nd law
we know that
it is product of mass and acceleration
here we know that
also we know that
so from above equation we have
so the force here will be 10020 N