Answer:
(a) 0.42 m
(b) 20.16 N/m
(c) - 0.42 m
(d) - 0.21 m
(e) 17.3 s
Solution:
As per the question:
Mass, m = 0.56 kg
x(t) = (0.42 m)cos[cos(6 rad/s)t]
Now,
The general eqn is:
![x(t) = Acos\omega t](https://tex.z-dn.net/?f=x%28t%29%20%3D%20Acos%5Comega%20t)
where
A = Amplitude
= angular frequency
t = time
Now, on comparing the given eqn with the general eqn:
(a) The amplitude of oscillation:
A = 0.42 m
(b) Spring constant k is given by:
![\omega = \sqrt{k}{m}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7Bk%7D%7Bm%7D)
![\omega^{2} = \frac{k}{m}](https://tex.z-dn.net/?f=%5Comega%5E%7B2%7D%20%3D%20%5Cfrac%7Bk%7D%7Bm%7D)
Thus
![k = m\omega^{2} = 0.56\times 6^{2} = 20.16\ N/m](https://tex.z-dn.net/?f=k%20%3D%20m%5Comega%5E%7B2%7D%20%3D%200.56%5Ctimes%206%5E%7B2%7D%20%3D%2020.16%5C%20N%2Fm)
(c) Position after one half period:
![x(t) = 0.42cos\pi = - 0.42\ m](https://tex.z-dn.net/?f=x%28t%29%20%3D%200.42cos%5Cpi%20%3D%20-%200.42%5C%20m)
(d) After one third of the period:
![x(t) = 0.42cos(\frac{2\pi}{3}) = - 0.21\ m](https://tex.z-dn.net/?f=x%28t%29%20%3D%200.42cos%28%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%20%3D%20-%200.21%5C%20m)
(e) Time taken to get at x = - 0.10 m:
![-0.10 = 0.42cos6t](https://tex.z-dn.net/?f=-0.10%20%3D%200.42cos6t)
![6t = co^{- 1} \frac{- 0.10}{0.42}](https://tex.z-dn.net/?f=6t%20%3D%20co%5E%7B-%201%7D%20%5Cfrac%7B-%200.10%7D%7B0.42%7D)
t = 17.3 s
I’m unsure whether there is a question there but in forces the force weight which includes gravity would be pushing the diver directly down. However if there was enough thrust from the diver the motion would change but ultimately will be accelerating downwards because of unbalanced forces
The final velocity of the other student after the elastic collision with Logan is 6.94 m/s.
<h3>
Conservation of linear momentum</h3>
The final velocity of the other student will be determined by applying the principle of conservation of linear momentum for elastic collision.
- let u represent initial velocity
- let v represent final velocity
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Substitute the given parameters and solve for final velocity of the other stsudent.
74(7.6) + 81(0) = 74(0) + 81(v₂)
562.4 = 81v₂
v₂ = 562.4/81
v₂ = 6.94 m/s
Thus, the final velocity of the other student after the elastic collision with Logan is 6.94 m/s.
Learn more about conservation of linear momentum here: brainly.com/question/7538238
![2.6×10^6\:\text{m}](https://tex.z-dn.net/?f=2.6%C3%9710%5E6%5C%3A%5Ctext%7Bm%7D)
Explanation:
The acceleration due to gravity g is defined as
![g = G\dfrac{M}{R^2}](https://tex.z-dn.net/?f=g%20%3D%20G%5Cdfrac%7BM%7D%7BR%5E2%7D)
and solving for R, we find that
![R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B%5Cdfrac%7BGM%7D%7Bg%7D%7D%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%281%29)
We need the mass M of the planet first and we can do that by noting that the centripetal acceleration
experienced by the satellite is equal to the gravitational force
or
![F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)](https://tex.z-dn.net/?f=F_c%20%3D%20F_G%20%5CRightarrow%20m%5Cdfrac%7Bv%5E2%7D%7Br%7D%20%3D%20G%5Cdfrac%7BmM%7D%7Br%5E2%7D%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%282%29)
The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as
![v = \dfrac{2\pi r}{T}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7B2%5Cpi%20r%7D%7BT%7D)
where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as
![\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Cpi%5E2%20r%7D%7BT%5E2%7D%20%3D%20G%5Cdfrac%7BM%7D%7Br%5E2%7D)
Solving for <em>M</em>, we get
![M = \dfrac{4\pi^2 r^3}{GT^2}](https://tex.z-dn.net/?f=M%20%3D%20%5Cdfrac%7B4%5Cpi%5E2%20r%5E3%7D%7BGT%5E2%7D)
Putting this expression back into Eqn(1), we get
![R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B%5Cdfrac%7BG%7D%7Bg%7D%5Cleft%28%5Cdfrac%7B4%5Cpi%5E2%20r%5E3%7D%7BGT%5E2%7D%5Cright%29%7D)
![\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3A%3D%5Cdfrac%7B2%5Cpi%7D%7BT%7D%5Csqrt%7B%5Cdfrac%7Br%5E3%7D%7Bg%7D%7D)
![\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3A%3D%5Cdfrac%7B2%5Cpi%7D%7B%281.44%C3%9710%5E4%5C%3A%5Ctext%7Bs%7D%29%7D%5Csqrt%7B%5Cdfrac%7B%285%C3%9710%5E6%5C%3A%5Ctext%7Bm%7D%29%5E3%7D%7B%283.45%5C%3A%5Ctext%7Bm%2Fs%7D%5E2%29%7D%7D)
![\:\:\:\:= 2.6×10^6\:\text{m}](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3A%3D%202.6%C3%9710%5E6%5C%3A%5Ctext%7Bm%7D)
The answer is: Diameter = d = 6.1 cm
Explanation:
Angular separation between the two stars = Q = 10^-5 rad.
Wavelength =λ= 500 * 10^-9 m
The smallest diameter = d = ?
As Q = 1.22*(λ/ d)
From this
d = 1.22*(λ/Q) --- (1)
Plug in the value in (1):
(1) => d = 1.22*((500*10^-9)/10^-5))
d = 0.061 m
d = 6.1 cm