Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.
The x location therefore is 290*cos(53)*35 = 6108.4m
The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.
This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m
So your (x,y) coordinates equals (6108.4, 2097.5)
Answer:
The atoms are aligned in a particular direction
Explanation:
The atoms become aligned in a particular direction in regions called domains, thus resulting in an overall resultant magnetism due to the spin of the electrons.
Answer:
The tunnel probability for 0.5 nm and 1.00 nm are 
and 
respectively.
Explanation:
Given that,
Energy E = 2 eV
Barrier V₀= 5.0 eV
Width = 1.00 nm
We need to calculate the value of 
Using formula of
Put the value into the formula
(a). We need to calculate the tunnel probability for width 0.5 nm
Using formula of tunnel barrier
Put the value into the formula
(b). We need to calculate the tunnel probability for width 1.00 nm
Hence, The tunnel probability for 0.5 nm and 1.00 nm are and respectively.
Answer:
the reflected wavelength is lano = 4.55 10⁻⁷ m which corresponds to the blue color
Explanation:
This is a case of reflection interference, we must be careful
* There is a 180º phase change when light passes from the air to the soap film (n = 1,339), but there is no phase change when passing from the pomp to the plastic (n = 1.3)
* the wavelength within the film is modulated by the refractive index
λₙ = λ₀ / n
if we consider these relationships the condition for constructive interference is
2 t = (m + ½) λₙ
2t = (m + ½) λ₀ / n
λ₀ = 2t n / (m + ½)
we substitute the values
λ₀= 2 255 10⁻⁹ 1,339 / (m + ½)
λ₀ = 6.829 10⁻⁷ (m + ½)
let's calculate the wavelength for various interference orders
m = 0
λ₀ = 6.829 10⁻⁷/ ( 0 + ½ )
λ₀ = 13.6 10⁻⁷
it is not visible
m = 1
λ₀ = 6,829 10⁻⁷/ (1 + ½)
λ₀ = 4.55 10⁻⁷
color blue
m = 2
λ₀ = 6.829 10⁻⁷ / (2 + ½)
λ₀ = 2,7 10⁻⁷
it is not visible
therefore the reflected wavelength is lano = 4.55 10⁻⁷ m which corresponds to the blue color
