To find the surface area, you need to do the following:
A side of the figure is a trapezoid, so you should do 13+5x7 (since there are 2 of the same sides there is no need to divide by two.) and it should equal 126. To find the area of the top of the figure, you need to do 5x2 which equals 10. The area of the bottom of the figure is 26 (13x2), then do 2x3x2 to get the sides on the left and right of the figure. After all of this, add them together. (126+10+26+12=174.)
The surface area should be 174mm.
Answer:
Answer:
x = 3.3
EF = 31.1
FG = 21.7
Step-by-step explanation:
Step-by-step explanation:
18 that question is confusing so I guessed sorry dude
The acceleration of the particle is given by the formula mentioned below:
Differentiate the position vector with respect to t.
![\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bds%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Csqrt%5B%5D%7B%5Cmleft%28t%5E3%2B1%5Cmright%29%7D%20%5C%5C%20%3D-%5Cfrac%7B1%7D%7B2%7D%28t%5E3%2B1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ctimes3t%5E2%20%5C%5C%20%3D%5Cfrac%7B3%7D%7B2%7D%5Cfrac%7Bt%5E2%7D%7B%5Csqrt%7B%28t%5E3%2B1%29%7D%7D%20%5Cend%7Bgathered%7D)
Differentiate both sides of the obtained equation with respect to t.
![\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bd%5E2s%28t%29%7D%7Bdx%5E2%7D%3D%5Cfrac%7B3%7D%7B2%7D%28%5Cfrac%7B2t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D%2Bt%5E2%28-%5Cfrac%7B3%7D%7B2%7D%29%5Ctimes%5Cfrac%7B1%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%29%20%5C%5C%20%3D%5Cfrac%7B3t%7D%7B%5Csqrt%5B%5D%7B%28t%5E3%2B1%29%7D%7D-%5Cfrac%7B9%7D%7B4%7D%5Cfrac%7Bt%5E2%7D%7B%28t%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5Cend%7Bgathered%7D)
Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.
![\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%28t%3D1%29%3D%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B2%7D%7D-%5Cfrac%7B9%7D%7B4%5Ctimes2%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%20%5C%5C%20%3D1.32ft%2Fsec%5E2%20%5Cend%7Bgathered%7D)
The initial position is obtained at t=0. Substitute t=0 in the given position function.
Answer:
0
Step-by-step explanation:
When you multiply any number by 0, the answer is always 0.
