D: Ultraviolet Rays is the answer hope that helps
Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration (
), measured in meters per square second, is estimated by this kinematic formula:
(1)
Where:
- Travelled distance, measured in meters.
, 
- Initial and final speeds of the spaceship, measured in meters.
If we know that 
, 
and 
, then the acceleration experimented by the spaceship is:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Given Information:
Diameter of spherical cell = 0.040 mm
thickness = L = 9 nm
Resistivity = ρ = 3.6×10⁷ Ω⋅m
Dielectric constant = k = 9.0
Required Information:
time constant = τ = ?
Answer:
time constant = 2.87×10⁻³ seconds
Explanation:
The time constant is given by
τ = RC
Where R is the resistance and C is the capacitance.
We know that resistivity of of any material is given by
ρ = RA/L
R = ρL/A
Where area of spherical cell is given by
A = 4πr²
A = 4π(d/2)²
A = 4π(0.040×10⁻³/2)²
A = 5.026×10⁻⁹ m²
The resistance becomes
R = (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹
R = 6.45×10⁷ Ω
The capacitance of the cell membrane is given by
C = kεoA/L
Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m
C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹
C = 44.5 pF
C = 44.5×10⁻¹² F
Therefore, the time constant is
τ = RC
τ = 6.45×10⁷*44.5×10⁻¹²
τ = 2.87×10⁻³ seconds
Scalar quantity are physical quantities that have just magnitude, not direction.
- It is always positive.
- Examples: Speed, distance
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
