V = u + at
<span>= 100 + 30*3 = 190 m/s </span>
<span>lets call the maximum altitude of the rocket H and say that the height reached while the engines are running is h. Also we'll call the angle of the launch a. </span>
<span>we can get an expression for h as follows. </span>
<span>h = vsin(a)t + (1/2)at^2 </span>
<span>h = 100*sin(53)*3 + 0.5*30*9 </span>
<span>h = 374.6 m </span>
<span>and we can write an expression for the time it takes the rocket to reach maximum height after the engines have failed. </span>
<span>t = [0 - vsin(a)]/-g </span>
<span>= vsin(a)/g </span>
<span>then </span>
<span>H-h = vsin(a)t - 1/2*gt^2 </span>
<span>if we sub in the expression for t and rearrange we get </span>
<span>H-h = [(vsin(a)^2]/2g </span>
<span>now add h </span>
<span>H = [(vsin(a)^2]/2g + h </span>
<span>H = 1549 m </span>
<span>b) </span>
<span>to obtain the total time of flight, first work out the time it takes for the rocket to fall from the top of its trajectory. </span>
<span>t = (v-u)/g </span>
<span>where v is now the final vertical velocity and u is the initial vertical velocity. </span>
<span>to get v </span>
<span>v^2 = u^2 + 2gH </span>
<span>so t = [Sqrt(2gH)]/g </span>
<span>now the total time is the falling time plus the time the engines are running, plus the time we calculated previously. </span>
<span>t = [Sqrt(2gH)]/g + 3 + vsin(a)/g </span>
<span>t = 36.72 s </span>
<span>c) </span>
<span>calculating the range is simple since the horizontal velocity stays the same for most of the flight. </span>
<span>We just need to work out the bit of distance traveled while the rocket accelerates for the first 3 seconds of flight. </span>
<span>r = ut + 1/2at^2 </span>
<span>r = 100*3+0.5*30*9 </span>
<span>r = 435 m </span>
<span>now we take the remaining flight time and put it in the following formula to work out the rest of the distance. </span>
<span>R-r = vcos(a)t </span>
<span>= 190*cos(53)*33.72 </span>
<span>= 3856 m </span>
<span>add this to the previous distance we calculated to obtain the range </span>
<span>R = 3856 + 435 </span>
<span>R = 4291 m </span>
<span>Hopefully that's right, I may have made an error in there so make sure you check through these calculations yourself to confirm. Hope this helps. :)</span>
This causes reverse faults<span>, which are the reverse of </span>normal faults<span>, because in this case, the hanging wall slides upward relative to the footwall. Shear </span>stress<span> is when rock slabs slide past each other horizontally. There is no vertical movement of either the hanging wall or footwall, and we get a strike-slip </span>fault<span>.</span>
if the color changes, it is neutral but if it stays the same, it is an acid.
Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to 
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.
As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be 
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
