Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
Answer:
The new voltage between the parallel plates of the capacitor is 18V, because for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage) between the plates.
Explanation:
ΔV = E*Δd
Where;
ΔV is the change in potential difference
Δd is the change in the distance between the parallel plates
E is the electric field potential.
Assuming a constant electric field; 
when the spacing between the capacitor plates is doubled, d₂ = 2d₁
v₂ = (v₁*d₂)/(d₁)
v₂ = (v₁*2d₁)/(d₁)
v₂ = 2v₁
v₂ = 2(9) = 18 V
Therefore, for a constant electric field, doubling the space between the parallel capacitor plates, will also double the potential difference (voltage).
I think the amplitude changed
Answer:
First 8 if wrong then correct me. Second it's direction correct me if wrong.
Explanation: