Answer:
<h2>18150 J</h2>
Explanation:
The kinetic energy of the car can be found by using the formula
![k = \frac{1}{2} m {v}^{2} \\](https://tex.z-dn.net/?f=k%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20m%20%7Bv%7D%5E%7B2%7D%20%20%5C%5C%20)
m is the Mass
v is the velocity
From the question we have
![k = \frac{1}{2} \times 1200 \times {5.5}^{2} \\ = 600 \times 30.25](https://tex.z-dn.net/?f=k%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%5Ctimes%201200%20%5Ctimes%20%20%7B5.5%7D%5E%7B2%7D%20%20%5C%5C%20%20%3D%20600%20%5Ctimes%2030.25)
We have the final answer as
<h3>18150 J</h3>
Hope this helps you
Answer:
1. The sound waves are longitudinal because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves.
2. A pulse or a wave is introduced into a slinky when a person holds the first coil and gives it a back-and-forth motion. This creates a disturbance within the medium; this disturbance subsequently travels from coil to coil, transporting energy as it moves.
Explanation:
Answer:
![q=1.95*10^{-7}C](https://tex.z-dn.net/?f=q%3D1.95%2A10%5E%7B-7%7DC)
Explanation:
According to the free-body diagram of the system, we have:
![\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)](https://tex.z-dn.net/?f=%5Csum%20F_y%3A%20Tcos%2815%5E%5Ccirc%29-mg%3D0%281%29%5C%5C%5Csum%20F_x%3A%20Tsin%2815%5E%5Ccirc%29-F_e%3D0%282%29)
So, we can solve for T from (1):
![T=\frac{mg}{cos(15^\circ)}(3)](https://tex.z-dn.net/?f=T%3D%5Cfrac%7Bmg%7D%7Bcos%2815%5E%5Ccirc%29%7D%283%29)
Replacing (3) in (2):
![(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e](https://tex.z-dn.net/?f=%28%5Cfrac%7Bmg%7D%7Bcos%2815%5E%5Ccirc%29%7D%29sin%2815%5E%5Ccirc%29%3DF_e)
The electric force (
) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:
![(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)](https://tex.z-dn.net/?f=%28%5Cfrac%7Bmg%7D%7Bcos%2815%5E%5Ccirc%29%7D%29sin%2815%5E%5Ccirc%29%3D%5Cfrac%7Bkq%5E2%7D%7Br%5E2%7D%284%29)
According to pythagoras theorem, the distance of separation (r) of the spheres are given by:
![sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)](https://tex.z-dn.net/?f=sin%2815%5E%5Ccirc%29%3D%5Cfrac%7B%5Cfrac%7Br%7D%7B2%7D%7D%7BL%7D%5C%5Cr%3D2Lsin%2815%5E%5Ccirc%29%285%29)
Finally, we replace (5) in (4) and solving for q:
![mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C](https://tex.z-dn.net/?f=mgtan%2815%5E%5Ccirc%29%3D%5Cfrac%7Bkq%5E2%7D%7B%282Lsin%2815%5E%5Ccirc%29%29%5E2%7D%5C%5Cq%3D%5Csqrt%7B%5Cfrac%7Bmgtan%2815%5E%5Ccirc%29%282Lsin%2815%5E%5Ccirc%29%29%5E2%7D%7Bk%7D%7D%5C%5Cq%3D%5Csqrt%7B%5Cfrac%7B10%2A10%5E%7B-3%7Dkg%289.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%29tan%2815%5E%5Ccirc%29%282%280.22m%29sin%2815%5E%5Ccirc%29%29%5E2%7D%7B8.98%2A10%5E%7B9%7D%5Cfrac%7BN%5Ccdot%20m%5E2%7D%7BC%5E2%7D%7D%7D%5C%5Cq%3D1.95%2A10%5E%7B-7%7DC)
Given parameters:
Initial velocity of Coin = 0m/s
Time taken before coin hits ground = 5.7s
Unknown:
Final velocity of the coin = ?
Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.
The fitting one of them here is shown below;
V = U + gt
where;
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
t is the time taken
Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.
Now input the parameters and solve;
V = 0 + 9.81 x 5.7
V = 55.917m/s
Therefore, the final velocity is 55.917m/s.