Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × 
................2
put here value
period of oscillations = 2π × 
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Answer:
T = 3.23 s
Explanation:
In the simple harmonic movement of a spring with a mass the angular velocity is given by
w = √ K / m
With the initial data let's look for the ratio k / m
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
2π / T = √ k / m
k₀ / m₀ = (2π / T)²
k₀ / m₀ = (2π / 3.0)²
k₀ / m₀ = 4.3865
The period on the new planet is
2π / T = √ k / m
T = 2π √ m / k
In this case the amounts are
m = 6 m₀
k = 10 k₀
We replace
T = 2π√6m₀ / 10k₀
T = 2π √6/10 √m₀ / k₀
T = 2π √ 0.6 √1 / 4.3865
T = 3.23 s
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Answer:
the pressure at the depth is 1.08 × 
Pa
Explanation:
The pressure at the depth is given by,
P = h 
g
Where, P = pressure at the depth
h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter
= density of water = 1000 
g = acceleration due to gravity ≈ 9.8 
P = 11033.15 × 9.8 × 1000
P = 1.08 × Pa
Thus, the pressure at the depth is 1.08 × Pa
