Incandescence is light at high temp
Answer:
891 excess electrons must be present on each sphere
Explanation:
One Charge = q1 = q
Force = F = 4.57*10^-21 N
Other charge = q2 =q
Distance = r = 20 cm = 0.2 m
permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)
Using Coulomb's law,
F=[1/4pieo]q1q2/r^2
F = [1/4pieo]q^2 / r^2
q^2 =F [4pieo]r^2
q = r*sq rt F[4pieo]
q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]
q = 1.42614*10^ -16 C
number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19
n =891
891 excess electrons must be present on each sphere
Answer:
atoms atoms atoms atoms atoms
Answer:
a) ΔV = 2,118 10⁻⁸ m³ b) ΔR= 0.0143 cm
Explanation:
a) For this part we use the concept of density
ρ = m / V
As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats
m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg
V = m / ρ
V = 0.56 / 3.52
V = 0.159 cm3
We use the relation of the bulk module
B = P / (Δv/V)
ΔV = V P / B
ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹
ΔV = 2,118 10⁻⁸ m³
b) indicates that we approximate the diamond to a sphere
V = 4/3 π R³
For this part let's look for the initial radius
R₀ = ∛ ¾ V /π
R₀ = ∛ (¾ 0.159 /π)
R₀ = 0.3361 cm
Now we look for the final volume and with this the final radius
= V + ΔV
= 0.159 + 2.118 10⁻²
= 0.18018 cm3
= ∛ (¾ 0.18018 /π)
= 0.3504 cm
The radius increment is
ΔR =
- R₀
ΔR = 0.3504 - 0.3361
ΔR= 0.0143 cm