Using Placks’s constant and frequency
t=5s
it was correct on my do-now
so I hope it was useful for you
Answer:
for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)
It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.
we have D= D0exp( -Qd/RT)
=(8.5×105m2/s)exp(-202,100/8.31×1023)
= 4.03 ×10-15m2/s
Answer:
The horizontal component of the velocity vector is;
vh = 34.4 ft/s
The vertical component of the velocity vector is;
vy = 49.1 ft/s
Explanation:
Given;
Velocity of football v = 60 ft/s
Angle of elevation ∅ = 55°
The horizontal component of the velocity vector is;
vh = vcos∅
Substituting the values;
vh = 60cos55°
vh = 34.41458618106 ft/s
vh = 34.4 ft/s
The vertical component of the velocity vector is;
vy = vsin∅
Substituting the values;
vy = 60sin55°
vy = 49.14912265733 ft/s
vy = 49.1 ft/s
Answer:
r = 2,026 10⁹ m and T = 2.027 10⁴ s
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F =
Acceleration is centripetal
a = v² / r
we substitute
r =
(1)
let's look for the charge in the insulating sphere
ρ = q₂ / V
q₂ = ρ V
the volume of the sphere is
v = 4/3 π r³
we substitute
q₂ = ρ
π r³
q₂ = 3 10⁻⁹
π 4³
q₂ = 8.04 10⁻⁷ C
let's calculate the radius with equation 1
r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)
r = 2,026 10⁹ m
this is the radius of the electron orbit around the charged sphere.
Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio
v = x / t
the distance traveled in a circle is
x = 2π r
In this case, time is the period
v = 2π r /T
T = 2π r /v
let's calculate
T = 2π 2,026 10⁹/628 103
T = 2.027 10⁴ s