Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π 
f= 1/2π 
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz
Answer:
a) I = 3.63 W / m²
, b) I = 0.750 W / m²
Explanation:
The intensity of a sound wave is given by the relation
I = P / A = ½ ρ v (2π f 
)²
I = (½ ρ v 4π² s_{max}²) f²
a) with the initial condition let's call the intensity Io
cte = (½ ρ v 4π² s_{max}²)
I₀ = cte s² f₀²
I₀ = cte 10 6
If frequency is increase f = 2.20 10³ Hz
I = constant (2.20 10³) 2
I = cte 4.84 10⁶
let's find the relationship of the two quantities
I / Io = 4.84
I = 4.84 Io
I = 4.84 0.750
I = 3.63 W / m²
b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or
I = cte (f s)²
I = constant (0.250 10³ 4)²
I = cte 1 10⁶
the relationship
I / Io = 1
I = Io
I = 0.750 W / m²
Answer: W = 
J
Explanation: Since the potassium ion is at the outside membrane of a cell and the potential here is lower than the potential inside the cell, the transport will need work to happen.
The work to transport an ion from a lower potential side to a higher potential side is calculated by
q is charge;
ΔV is the potential difference;
Potassium ion has +1 charge, which means:
p = 
C
To determine work in joules, potential has to be in Volts, so:
Then, work is
To move a potassium ion from the exterior to the interior of the cell, it is required J of energy.
