Answer:
The displacement in t = 0,
y (0) = - 0.18 m
Explanation:
Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N
v = √ T / μ
v = √20.48 N / 0.02 kg /m = 32 m/s
λ = v / f
λ = 32 m/s / 40 Hz = 0.8
K = 2 π / λ
K = 2π / 0.8 = 7.854
φ = X * 360 / λ
φ = 0.5 * 360 / 0.8 = 225 °
Using the model of y' displacement
y (t) = A* sin ( w * t - φ )
When t = 0
y (0) = 0.25 m *sin ( w*(0) - 225 )
y (0) = 0.25 * -0.707
y (0) = - 0.18 m
Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is
Explanation:
From the question we are told that
The semi - major axis of the rocky debris 
The semi - major axis of Planet D is 
The orbital period of planet D is 
Generally from Kepler third law

Here T is the orbital period while a is the semi major axis
So

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=> ![T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%2018.164%20%20%2A%20%20%5B%5Cfrac%7B%2045%7D%7B60%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
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Answer:
ow much work was done? W = F xD. IN X 2m = 2;. 2. A force of 15 newtons is ... 3. It took 50 joules to push a chair 5 meters across the floor. With what force was ... was done. How far was the rock lifted? W=FXD. D=1500 = 1.5m. Answer: :.5m ... A young man exerted a force of 9,000 newtons on a stalled car, but he was.
Explanation:
We will apply the concept of period in a pendulum, defined as the product between 2
by the square root of the length over gravity, this is mathematically

Here,
T = Period
L = Length
g = Acceleration due to gravity
For the period to be 1 second, then we must look for the necessary length for such a requirement so




The meter's length would be slight less than one-fourth of its current length. Also, the number of significant digits depends only on how precisely we know g, because the time has been defined to be exactly 1s.
Therefore the correct answer is C.