Answer:
t = 6 [s]
Explanation:
In order to solve this problem we must first use this equation of kinematics.
![v_{f}^{2}=v_{o}^{2} +2*a*x](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3Dv_%7Bo%7D%5E%7B2%7D%20%2B2%2Aa%2Ax)
where:
Vf = final velocity = 0 (the car comes to rest)
Vo = initial velocity = 72 [km/h]
a = acceleration [m/s²]
x = distance = 60 [m]
First we must convert the velocity from kilometers per hour to meters per second.
![72 [\frac{km}{h}]*\frac{1000m}{1km} *\frac{1h}{3600s} =20 [m/s]](https://tex.z-dn.net/?f=72%20%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A%5Cfrac%7B1000m%7D%7B1km%7D%20%2A%5Cfrac%7B1h%7D%7B3600s%7D%20%3D20%20%5Bm%2Fs%5D)
![0=(20)^{2} -2*a*60\\400 = 120*a\\a=3.33[m/s^{2} ]](https://tex.z-dn.net/?f=0%3D%2820%29%5E%7B2%7D%20-2%2Aa%2A60%5C%5C400%20%3D%20120%2Aa%5C%5Ca%3D3.33%5Bm%2Fs%5E%7B2%7D%20%5D)
Now using this other equation of kinematics.
![v_{f}=v_{o}-a*t](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At)
0 = 20-3.33*t
t = 6[s]
To me the basic difference is their frequency and wavelength.
For frequency :
Infrared < visible light < ultraviolet
For wavelength :
Infrared > visible light > ultraviolet
Answer:
Well..
Explanation:
That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.
Answer:
The image of everything in front of the mirror is reflected backward, retracing the path it traveled to get there. Nothing is switching left to right or up-down. Instead, it's being inverted front to back. ... That reflection represents the photons of light, bouncing back in the same direction from which they came
Explanation:
Answer:
1. the graph is showing the speed of the object at different times.
Explanation:
2. the object is slowing down