Answer:
magnitude of the pressure on the sail in the vicinity of Earth’s Orbit=
Explanation:
The momentum of a photon is:
p = E/c
E = the photon energy
c = the speed of light.
take the time derivative (gives the force)
F = dp/dt = (dE/dt)/c
F = 2(dE/dt)/c (is doubled for complete reflection of the light)
Intensity has the units of energy per unit time per unit area
= I
then,
Force/unit area = 2I/c
magnitude of the pressure on the sail in the vicinity of Earth’s Orbit=
Answer:
ac= 15.07 m/s²
Explanation:
The Wheel rotates with a constant angular acceleration.:
Centripetal acceleration is calculated as follows:
ac =ω² *R Formula (1)
ac =v² / R Formula (2)
Kinematics of the wheel
We apply the equations of circular motion uniformly accelerated :
ωf²= ω₀² + 2αθ Formula (3)
v = ω* R Formula (4)
Where:
θ : angle that the wheel has rotated in a given time interval (rad)
α : angular acceleration (rad/s²)
ω₀ : initial angular speed ( rad/s)
ωf : final angular speed ( rad/s)
v: tangential velocity of a point on the rim ( m/s)
R : radius of wheel (m)
ac: centripetal acceleration, (m/s²)
Data:
D = 40.0 cm : diameter of the wheel
R = D/2= 40.0 cm/ 2 = 20 cm = 0.2m
α = 3.00 rad/s^2
ω₀ = 0
n = 2 revolutions : number of revolutions
θ =2πn (rad) = 2π*2 (rad) = 4π rad
Calculate of the ωf
We replace data in the formula (3)
ωf²= ω₀² + 2αθ
ωf²= 0 + 2(3)(4π)
ωf²= 24π
ωf = 8.68 rad/s
Calculate of the v
We replace data in the formula (4)
v = ω*R
v = (8.68)*(0.2)
v = 1.736 m/s
Calculate of the ac
We replace data in the formula (1)
ac = ( ω)²*(R)
ac = (8.68)²*(0.2)
ac = 15.07 m/s²
We replace data in the formula (2)
ac = v²/ R
ac = ( 1.736 )²/(0.2)
ac = 15.07 m/s²
To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity
At constant temperature and pressure, the change in Gibbs free energy is defined as
Where,
H = Entalpy
T = Temperature
S = Entropy
When the temperature is less than that number it is negative meaning it is a spontaneous reaction. is also always 0 when using single element reactions. In numerical that implies
At the equation then,
}
Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K