Just about anything you could ask about a projectile AFTER it's launched
depends on both components of the launch velocity.
Here are some that I can think of:
-- angle of launch
-- magnitude of launch velocity
-- location at any time after launch
-- magnitude of velocity at any time after launch
-- direction of velocity at any time after launch
-- distance of the landing point from the launch point
To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.
By Hook's law we know that force is defined as,
![F= kx](https://tex.z-dn.net/?f=F%3D%20kx)
Where,
k = spring constant
x = Displacement change
PART A) For the case of the spring constant we can use the above equation and clear k so that
![k= \frac{F}{x}](https://tex.z-dn.net/?f=k%3D%20%5Cfrac%7BF%7D%7Bx%7D)
![k = \frac{mg}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7Bmg%7D%7Bx%7D)
![k= \frac{77.2*9.8}{0.0637}](https://tex.z-dn.net/?f=k%3D%20%5Cfrac%7B77.2%2A9.8%7D%7B0.0637%7D)
![k = 11876.92N/m](https://tex.z-dn.net/?f=k%20%3D%2011876.92N%2Fm)
Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m
PART B) In the case of speed we can obtain it through the period, which is given by
![T = \frac{2\pi}{\omega}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Comega%7D)
Re-arrange to find \omega,
![\omega = \frac{2\pi}{T}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B2%5Cpi%7D%7BT%7D)
![\omega = \frac{2\pi}{2.14}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B2.14%7D)
![\omega = 2.93rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202.93rad%2Fs)
Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to
![\omega^2 = \frac{k}{m}](https://tex.z-dn.net/?f=%5Comega%5E2%20%3D%20%5Cfrac%7Bk%7D%7Bm%7D)
![m = \frac{k}{\omega^2}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bk%7D%7B%5Comega%5E2%7D)
![m = \frac{ 11876.92}{2.93}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B%2011876.92%7D%7B2.93%7D)
![m = 4093.55Kg](https://tex.z-dn.net/?f=m%20%3D%204093.55Kg)
Therefore the mass of the trailer is 4093.55Kg
PART C) The frequency by definition is inversely to the period therefore
![f = \frac{1}{T}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7BT%7D)
![f = \frac{1}{2.14}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B1%7D%7B2.14%7D)
![f = 0.4672 Hz](https://tex.z-dn.net/?f=f%20%3D%200.4672%20Hz)
Therefore the frequency of the oscillation is 0.4672 Hz
PART D) The time it takes to make the route 10 times would be 10 times the period, that is
![t_T = 10*T](https://tex.z-dn.net/?f=t_T%20%3D%2010%2AT)
![t_T = 10 *2.14s](https://tex.z-dn.net/?f=t_T%20%3D%2010%20%2A2.14s)
![t_T = 21.4s](https://tex.z-dn.net/?f=t_T%20%3D%2021.4s)
Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s
Answer
Explanation:
As the three resistors are connected in series, the expression to be used for the
calculation of RT equivalent resistance
is:
RT = R1 + R2 + R3
We replace the data of the statement in the previous expression and it remains:
5 10 15 RT + R1 + R2 + R3 + +
We perform the mathematical operations that lead us to the result we are looking for:
RT - 30Ω
The spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action