Answer:
1 P = 0.5
2 P = 0.3
3 P = 0.01
Explanation:
The probability formula is
Where P is the probability V is the volume while N is the number of distinguishing particles
So for N = 1 and 
= 0.5
For N = 1 and 
= 0.3
For N = 1 and 
= 0.01
Answer:
a) v = 2,9992 10⁸ m / s
, b) Eo = 375 V / m
, B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz
, T = 1.05 10⁻¹⁵ s
, UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
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Explanation:
When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017
