Well I don't know !
Let's work it out.
The gravitational force between two objects is
F = G · M₁·M₂ / R² .
'G' is the 'universal gravitational constant'. We could look it up.
'M₁' is the mass of one object
'M₂' is the mass of the other object
'R' is the distance between their centers.
It looks complicated, but stay with me. We can do this !
We know all the numbers, so we can calculate the force.
'G' is 6.67 x 10⁻¹¹ newton·meter² / kg² (I looked it up. You're welcome.)
'M₁' is 15 kg
'M₂' is 15 kg
'R' is 0.25 meter.
Now it's time to pluggum in.
F = G · M₁·M₂ / R²
= (6.67 x 10⁻¹¹ newton·meter² / kg²) · (15 kg) · (15 kg) / (0.25 m²)
= (6.67 x 10⁻¹¹ · 15 · 15 / 0.0625) N·m²·kg·kg / kg²·m²
= 2.4 x 10⁻⁷ Newton .
That a force equivalent to about 0.00000086 of an ounce.
This is the answer to part-a.
Concerning the answer to part-b ...
Personally, I could not detect this force, no matter what kind of equipment
I had. But I am just a poor schlepper engineer, educated in the last Century,
living out my days on Brainly and getting my kicks from YouTube videos.
I am not pushing the box to the envelope, or thinking outside the cutting
edge ... whatever.
I am sure there are people ... I can't name them, because they keep a
low profile, they stay under the radar, they don't attract a lot of media
attention, their work is not as newsworthy as the Kardashians, and plus,
they seldom call me or write to me ... but I know in my bones that there
are people who have measured the speed of light to NINE significant figures,
aimed a spacecraft accurately enough to take close-up pix of Pluto ten years
later, and detected gravity waves from massive blobs that merged 13 billion
years ago, and I tell you that YES ! THESE guys could detect and measure
a force of 0.86 micro-ounce if they felt like it !
The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
First we need to convert the frequency to the correct SI units (seconds)
Frequency of bell = 1440/60 = 24 Hertz (Strikes per second)
Power is the rate of work being done or the work done per second.
P = E/t = (24 * 0.2)/1 = 4.8 Watt
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