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
I believe that the answer is <span>C. The atmosphere scatters blue light more than yellow light.</span>
Answer:
1.2 m/s
0.31 m
0.15 m
Explanation:
Time period is
Frequency is
Velocity is given by
The waves are traveling at 1.2 m/s
Amplitude is given by
Amplitude is 0.31 m
If d = 0.3 m
The amplitude would be 0.15 m. The speed would remain the same.
The velocity of pluck 1 is 12 m/s west.
<h3>What is the conservation of momentum?</h3>
The principle of the conservation of the linear momentum states that momentum before collision is equal to momentum after collision.
Now given that;
m1u1 + m2u2 = m1v1 + m2v2
(0.1 * 15) - (0.1 * 12) = 0.1* v + (0.1 * 15)
1.5 - 1.2 = 0.1v + 1.5
0.3 - 1.5 = 0.1v
v = -1.2/0.1
v = - 12 m/s
Hence, the velocity of pluck 1 is 12 m/s west.
Learn more about linear momentum:brainly.com/question/27988315
#SPJ1
Answer:
o to increase the frequency of sound waves. It increases the sound waves to a level of frequency that humans cannot hear so you won't be able to hear many things though the wall other then low noises like pounding.
Explanation:
I am in construction class as well as a student teacher for other construction type programs trust me :D
Brainiest would be appreciated
