Answer:
a) ![a=4\,\frac{m}{s^2}](https://tex.z-dn.net/?f=a%3D4%5C%2C%5Cfrac%7Bm%7D%7Bs%5E2%7D)
b) ![V(t)=4\,t\,+3](https://tex.z-dn.net/?f=V%28t%29%3D4%5C%2Ct%5C%2C%2B3)
c) ![V(1)=7 \,\frac{m}{s} \\](https://tex.z-dn.net/?f=V%281%29%3D7%20%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5C)
d) Displacement = 22 m
e) Average speed = 11 m/s
Explanation:
a)
Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:
![slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}](https://tex.z-dn.net/?f=slope%3D%20%5Cfrac%7B15%3D3%7D%7B3-0%7D%20%3D4%5C%2C%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Therefore, acceleration is ![a=4\,\frac{m}{s^2}](https://tex.z-dn.net/?f=a%3D4%5C%2C%5Cfrac%7Bm%7D%7Bs%5E2%7D)
b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:
![y=m\,x+b\\V(t)=4\,t\,+3](https://tex.z-dn.net/?f=y%3Dm%5C%2Cx%2Bb%5C%5CV%28t%29%3D4%5C%2Ct%5C%2C%2B3)
c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:
![V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}](https://tex.z-dn.net/?f=V%28t%29%3D%204%5C%2Ct%2B3%5C%5CV%281%29%3D4%5C%2C%281%29%2B3%5C%5CV%281%29%3D7%20%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):
Displacement = ![\frac{(7+15)\,2}{2} = 22\,\,m](https://tex.z-dn.net/?f=%5Cfrac%7B%287%2B15%29%5C%2C2%7D%7B2%7D%20%3D%2022%5C%2C%5C%2Cm)
e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:
Average velocity = ![\frac{22}{2} = 11\, \,\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7B22%7D%7B2%7D%20%3D%2011%5C%2C%20%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Answer:
for example the studs are made in football player boot because to increase pressure with descrease in area of surface
Answer:
Option D: 1.5in in front of the target
Explanation:
The object distance is
.
Because the surface is flat, the radius of curvature is infinity .
The incident index is
and the transmitted index is
.
The single interface equation is ![\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_i%7D%7By%7D%2B%5Cfrac%7Bn_t%7D%7By%5Ei%7D%3D%5Cfrac%7Bn_t-n_i%7D%7Br%7D)
Substituting the quantities given in the problem,
![\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B4%7D%7B3%7D%7D%7B6in%7D%2B%5Cfrac%7B1%7D%7By%5Ei%7D%3D%200)
The image distance is then ![y^i=-\frac{18}{4}in =-4.5in](https://tex.z-dn.net/?f=y%5Ei%3D-%5Cfrac%7B18%7D%7B4%7Din%20%3D-4.5in)
Therefore, the coin falls
in front of the target
B, a disurbance would make it unstable and affect the center of mass, which would then affect the equilibrium.