Answer:
Each layer has its own properties, composition, and characteristics that affects many of the key processes of our planet. They are, in order from the exterior to the interior – the crust, the mantle, the outer core, and the inner core.
Explanation:
Electromagnetic waves<span> transfer energy without going through a medium. ... Sometimes, a </span>transverse wave<span> and a </span>longitudinal wave can combine to form<span>another </span>kind<span> of </span>wave<span> called a surface </span>wave<span>. </span>Transverse Waves<span>. </span>Waves<span> in which the particles vibrate in an up-and-down motion
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Answer:
Though there is no chart on my screen, I can give you the correct information to solve the question without knowing what it looks like. Assuming that this track has slopes and or ramps of any kind, the place where it would have the most potential energy would be at the highest point of this ramp, or and the highest point on the track. This would also mean that it would fall the farthest. The highest kinetic energy would be at the lowest part of the ramp, after its used the slope to its full ability. But this would also be before it hits flat ground, or starts going up another ramp. This would be the point where it would be going its fastest.
Explanation:
Answer:
t = 0.55[sg]; v = 0.9[m/s]
Explanation:
In order to solve this problem we must establish the initial conditions with which we can work.
y = initial elevation = - 1.5 [m]
x = landing distance = 0.5 [m]
We set "y" with a negative value, as this height is below the table level.
in the following equation (vy)o is equal to zero because there is no velocity in the y component.
therefore:
![y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7Bo%7D%2At%20-%20%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%5C%5C%20%20%20where%3A%5C%5C%28v_%7By%7D%29_%7Bo%7D%3D0%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%5Bsg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%5D%5C%5C%20-1.5%20%3D%200%2At%20-4.905%2At%5E%7B2%7D%20%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B1.5%7D%7B4.905%7D%20%7D%20%5C%5Ct%3D0.55%5Bs%5D)
Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.
![(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7B0.5%7D%7B0.55%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D0.9%5Bm%2Fs%5D)
