Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
Answer:
2649600 Joules
Explanation:
Efficiency = 40%
m = Mass of air = 92000 kg
v = Velocity of wind = 12 m/s
Kinetic energy is given by

The kinetic energy of the wind is 6624000 Joules
The wind turbine extracts 40% of the kinetic energy of the wind

The energy extracted by the turbine every second is 2649600 Joules
Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 ×
m
wavelength λ = 700 nm = 700 ×
m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is = 
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
sinθ = 1 × 380 × 
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
sinθ = 1 × 700 × 
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )
When you heat a certain substance with a difference of temperature

the heat (energy) you must give to it is

where

is the specific heat of that substance (given in J/(g*Celsius))
In this case

Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin) because on the temperature scale a
difference of 1 degree Celsius = 1 degree Kelvin
1. 2+0.5+2.5= 3. 2km/hr average
2. 14-6=4seconds. 8m/s in 4s = 2m/s acceleration
3. 15m/s divided by 2.5 = 6m/s acceleration