The newton is the SI unit for force; it is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared.
Answer:
![\frac{1}{8} y'' + 2y' + 24y=0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B8%7D%20y%27%27%20%2B%202y%27%20%2B%2024y%3D0)
Explanation:
The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by
![my'' + \zeta y' + ky=0](https://tex.z-dn.net/?f=my%27%27%20%2B%20%5Czeta%20y%27%20%2B%20ky%3D0)
Where m is the mass, ζ is the damping constant, and k is the spring constant.
The spring constant k can be found by
![w - kL=0](https://tex.z-dn.net/?f=w%20-%20kL%3D0)
![mg - kL=0](https://tex.z-dn.net/?f=mg%20-%20kL%3D0)
![4 - k\frac{1}{6}=0](https://tex.z-dn.net/?f=4%20-%20k%5Cfrac%7B1%7D%7B6%7D%3D0)
![k = 4\times 6 =24](https://tex.z-dn.net/?f=k%20%3D%204%5Ctimes%206%20%3D24)
The damping constant can be found by
![F = -\zeta y'](https://tex.z-dn.net/?f=F%20%3D%20-%5Czeta%20y%27)
![6 = 3\zeta](https://tex.z-dn.net/?f=6%20%3D%203%5Czeta)
![\zeta = \frac{6}{3} = 2](https://tex.z-dn.net/?f=%5Czeta%20%3D%20%5Cfrac%7B6%7D%7B3%7D%20%3D%202)
Finally, the mass m can be found by
![w = 4](https://tex.z-dn.net/?f=w%20%3D%204)
![mg=4](https://tex.z-dn.net/?f=mg%3D4)
![m = \frac{4}{g}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B4%7D%7Bg%7D)
Where g is approximately 32 ft/s²
![m = \frac{4}{32} = \frac{1}{8}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B4%7D%7B32%7D%20%3D%20%5Cfrac%7B1%7D%7B8%7D)
Therefore, the required differential equation is
![my'' + \zeta y' + ky=0](https://tex.z-dn.net/?f=my%27%27%20%2B%20%5Czeta%20y%27%20%2B%20ky%3D0)
![\frac{1}{8} y'' + 2y' + 24y=0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B8%7D%20y%27%27%20%2B%202y%27%20%2B%2024y%3D0)
The initial position is
![y(0) = \frac{1}{2}](https://tex.z-dn.net/?f=y%280%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
The initial velocity is
![y'(0) = 0](https://tex.z-dn.net/?f=y%27%280%29%20%3D%200)
Answer:
0
Explanation:
It’s before the projectile was fired, so nothing has happened yet.
Answer:
Therefore it is save to carry a 62kg adult
Explanation:
From the question we are told that:
Mass ![m=380kg](https://tex.z-dn.net/?f=m%3D380kg)
Height of supporting Rock ![X=85cm](https://tex.z-dn.net/?f=X%3D85cm)
Length of Board![L_r=4.5m](https://tex.z-dn.net/?f=L_r%3D4.5m)
Mass of board ![M_b=22kg](https://tex.z-dn.net/?f=M_b%3D22kg)
Mass of adult
Generally the moment of balance about wedge part about is mathematically given by
![N -Q + R = Mg + mg](https://tex.z-dn.net/?f=N%20-Q%20%2B%20R%20%3D%20Mg%20%2B%20mg)
![0.85*N - Mg*2.25 - mg*(2.25 + x) = 0](https://tex.z-dn.net/?f=0.85%2AN%20-%20Mg%2A2.25%20-%20mg%2A%282.25%20%2B%20x%29%20%3D%200)
![0.85*N = + Mg*2.25 + mg*(2.25 + x)](https://tex.z-dn.net/?f=0.85%2AN%20%20%3D%20%2B%20Mg%2A2.25%20%2B%20mg%2A%282.25%20%2B%20x%29)
where
![N+R=4547](https://tex.z-dn.net/?f=N%2BR%3D4547)
therefore
![N = 570.70588 + 1608.3529 + 714.823 x](https://tex.z-dn.net/?f=N%20%3D%20570.70588%20%2B%201608.3529%20%2B%20714.823%20x)
if N=0 at fallen person
![x=3.04m](https://tex.z-dn.net/?f=x%3D3.04m)
Therefore it is save to carry a 62kg adult
Answer:
v = 3.08 km/s
Explanation:
Given that,
The angular velocity of the satellite = ![\omega=7.27\times 10^{-5} rad/s](https://tex.z-dn.net/?f=%5Comega%3D7.27%5Ctimes%2010%5E%7B-5%7D%20rad%2Fs)
A satellite is in orbit 36000km above the surface of the earth.
The radius of the earth is 6400 km
Let v is the velocity of the satellite. It can be calculated as :
![v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s](https://tex.z-dn.net/?f=v%3Dr%5Comega%5C%5C%5C%5Cv%3D%2836000%5Ctimes%2010%5E3%2B6400%5Ctimes%2010%5E3%29%5Ctimes%207.27%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cv%3D3082.48%5C%20m%2Fs%5C%5C%5C%5Cv%3D3.08%5C%20km%2Fs)
So, the velocity of the satellite is 3.08 km/s.