<span>A cinder cone is formed from relatively low viscosity magma with a high gas content. Cinder cones
are formed by the volcanic ash, cinders that are formed around the volcanic
vent. Cinder cones have a bowl shaped crater and the shape of the cinder cones
depend on the ejected particles that formed the cinder.</span>
|Acceleration| = (change in speed) / (time for the change).
Change in speed = (6 mi/hr - 25 mi/hr) = -19 mi/hr
Time for the change = 10 sec
|Acceleration| = (-19 mi/hr) / (10 sec) = -1.9 mile per hour per second
Admittedly, that's a rather weird unit.
Other units, perhaps more comfortable ones, are:
-6,840 mi/hr²
-2.79 feet/sec²
Remember, half of the energy in an EM wave is in the E field, the rest is in the B field.
Thus, multiply E field energy by 2.
To calculate the energy of the wave you must then use the following equation: W = A*t*c*2*(1/2*E^2*Eo). Where, A = Area, t = time, c = speed of light (which is a constant), E = Electric field, E0 = vacuum permittivity (8.85*10^-12 Nm^2/C^2). Substituting W =(0.320)*(26)*(3*10^8)*(2)*((1/2)*(1.95*10^-2)^2*(8.854*10^-12)) = 8.40*10^-6 J
Because of the gravitational pull when we go up the gravity pulls u down
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)
