Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C 
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
if there were no invention of machines then life would have been more difficult and simple works could be hard to do. Even now we are using our phones, sitting in a AC room interacting to eachother from different places. without the invention of machines simple things like transportation would have been difficult. There would be horses and donkey for the transportation. There would be no electricity,no internet, no transportation, not even c computers or mobile etc. The market for business will be smaller, the knowledge and news about world would be less.
so the problem would have been bigger than we can imagine. But one thing is that nature could survive lot more compared to what we have done till now by destroying nature.
Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.
Mass of the gold (m) = 0.072 kg
Temperature change (ΔT) = 90 - 20 = 70 degree Celsius
Specific heat capacity of the gold (c) = 136 J/kg C
Heat supplied = m × c × ΔT
Heat supplied = 0.072 × 136 × 70
Heat supplied = 685.44 Joules
Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules
Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,
K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J
Answer:
The constant angular acceleration of the centrifuge = -252.84 rad/s²
Explanation:
We will be using the equations of motion for this calculation.
Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.
First of, we write the given parameters.
w₀ = initial angular velocity = 2πf₀
f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s
w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s
θ = 46 revs = 46 × 2π = 289.14 rad
w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)
α = ?
Just like v² = u² + 2ay
w² = w₀² + 2αθ
0 = 382.38² + [2α × (289.14)]
578.29α = -146,214.4644
α = (-146,214.4644/578.29)
α = - 252.84 rad/s²
Hope this Helps!!!
