Answer:
15,872 mm³
Step-by-step explanation:
given:
A small square pyramid of height 6 cm was removed from the top of a large square pyramid of height 12cm forming the solid shown.
Find:
the exact volume of the solid
solution:
volume of square base pyramid = (base area)² * h/3
where total h = 12 cm
height of top pyramid (ht)= 6 cm
height of bottom pyramid (hb) = 6 cm
bottom volume = total volume - the volume on top
so,
total volume = 1/3 (base area)² h
= 1/3 (8*8)² * 12
= 16,384 mm³
volume on top = 1/3 (top base area)² h
= 1/3 (4*4)² * 6
= 512 mm³
finally: get the bottom volume:
bottom volume = total volume - the volume on top
bot. vol = 16,384 mm³ - 512 mm³
= 15,872 mm³
therefore,
the volume of the cut pyramid base = 15,872 mm³
Answer:
GPA stands for grade point average. the highest you can get is 4.5. There is no such things as a 95 GPA. To calculate, take the number or point per grade you have, A is 4.0, B is 3.0, C is 2.0, and so on, and add them all up and then divide by the number of classes you are taking.
Step-by-step explanation:
i did not understand the question can u explain again i can help u
Answer: 143 hamburgers and 429 cheese burgers
Explanation:
Call h and c the number of both items.
(h-hamburger and c-cheeseburger)
h + c = 572
c = 3h
Sub the second into the first
h + 3h = 572
4h = 572
Divide both sides by 4
h = 143 hamburgers
Use this back into the second equation
c = 3 • 143 = 429 cheeseburgers
Answer:
The curvature is ![\kappa=1](https://tex.z-dn.net/?f=%5Ckappa%3D1)
The tangential component of acceleration is ![a_{\boldsymbol{T}}=0](https://tex.z-dn.net/?f=a_%7B%5Cboldsymbol%7BT%7D%7D%3D0)
The normal component of acceleration is ![a_{\boldsymbol{N}}=1 (2)^2=4](https://tex.z-dn.net/?f=a_%7B%5Cboldsymbol%7BN%7D%7D%3D1%20%282%29%5E2%3D4)
Step-by-step explanation:
To find the curvature of the path we are going to use this formula:
![\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}](https://tex.z-dn.net/?f=%5Ckappa%3D%5Cfrac%7B%7C%7Cd%5Cboldsymbol%7BT%7D%2Fdt%7C%7C%7D%7Bds%2Fdt%7D)
where
is the unit tangent vector.
is the speed of the object
We need to find
, we know that
so
![\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}](https://tex.z-dn.net/?f=%5Cboldsymbol%7Br%7D%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%28cos%5Cleft%282t%5Cright%29%5Cright%29%5C%3A%5Cboldsymbol%7Bi%7D%2B%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%28sin%5Cleft%282t%5Cright%29%5Cright%29%5C%3A%5Cboldsymbol%7Bj%7D%2B%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%281%29%5Cright%5C%3A%5Cboldsymbol%7Bk%7D%5C%5C%5Cboldsymbol%7Br%7D%27%28t%29%3D-2%5Csin%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bi%7D%2B2%5Ccos%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bj%7D)
Next , we find the magnitude of derivative of the position vector
![|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2](https://tex.z-dn.net/?f=%7C%7C%20%5Cboldsymbol%7Br%7D%27%28t%29%7D%7C%7C%3D%5Csqrt%7B%28-2%5Csin%20%5Cleft%282t%5Cright%29%29%5E2%2B%282%5Ccos%20%5Cleft%282t%5Cright%29%29%5E2%7D%20%5C%5C%7C%7C%20%5Cboldsymbol%7Br%7D%27%28t%29%7D%7C%7C%3D%5Csqrt%7B2%5E2%5Csin%20%5E2%5Cleft%282t%5Cright%29%2B2%5E2%5Ccos%20%5E2%5Cleft%282t%5Cright%29%7D%5C%5C%7C%7C%20%5Cboldsymbol%7Br%7D%27%28t%29%7D%7C%7C%3D%5Csqrt%7B4%5Cleft%28%5Csin%20%5E2%5Cleft%282t%5Cright%29%2B%5Ccos%20%5E2%5Cleft%282t%5Cright%29%5Cright%29%7D%5C%5C%7C%7C%20%5Cboldsymbol%7Br%7D%27%28t%29%7D%7C%7C%3D%5Csqrt%7B4%7D%5Csqrt%7B%5Csin%20%5E2%5Cleft%282t%5Cright%29%2B%5Ccos%20%5E2%5Cleft%282t%5Cright%29%7D%5C%5C%5C%5C%5Cmathrm%7BUse%5C%3Athe%5C%3Afollowing%5C%3Aidentity%7D%3A%5Cquad%20%5Ccos%20%5E2%5Cleft%28x%5Cright%29%2B%5Csin%20%5E2%5Cleft%28x%5Cright%29%3D1%5C%5C%5C%5C%7C%7C%20%5Cboldsymbol%7Br%7D%27%28t%29%7D%7C%7C%3D2%5Csqrt%7B1%7D%3D2)
The unit tangent vector is defined by
![\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}](https://tex.z-dn.net/?f=%5Cboldsymbol%7BT%7D%7D%3D%5Cfrac%7B%5Cboldsymbol%7Br%7D%27%28t%29%7D%7B%7C%7C%5Cboldsymbol%7Br%7D%27%28t%29%7C%7C%7D)
![\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)](https://tex.z-dn.net/?f=%5Cboldsymbol%7BT%7D%7D%3D%5Cfrac%7B-2%5Csin%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bi%7D%2B2%5Ccos%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bj%7D%7D%7B2%7D%20%3D%5Csin%20%5Cleft%282t%5Cright%29%2B%5Ccos%20%5Cleft%282t%5Cright%29)
We need to find the derivative of unit tangent vector
![\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})](https://tex.z-dn.net/?f=%5Cboldsymbol%7BT%7D%27%3D%5Cfrac%7Bd%7D%7Bdt%7D%28%5Csin%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bi%7D%2B%5Ccos%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bj%7D%29%20%5C%5C%5Cboldsymbol%7BT%7D%27%3D-2%5Ccdot%28%5Csin%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bi%7D%2B%5Ccos%20%5Cleft%282t%5Cright%29%5Cboldsymbol%7Bj%7D%29)
And the magnitude of the derivative of unit tangent vector is
![||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2](https://tex.z-dn.net/?f=%7C%7C%5Cboldsymbol%7BT%7D%27%7C%7C%3D2%5Csqrt%7B%5Ccos%20%5E2%5Cleft%28x%5Cright%29%2B%5Csin%20%5E2%5Cleft%28x%5Cright%29%7D%20%3D2)
The curvature is
![\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1](https://tex.z-dn.net/?f=%5Ckappa%3D%5Cfrac%7B%7C%7Cd%5Cboldsymbol%7BT%7D%2Fdt%7C%7C%7D%7Bds%2Fdt%7D%3D%5Cfrac%7B2%7D%7B2%7D%20%3D1)
The tangential component of acceleration is given by the formula
![a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}](https://tex.z-dn.net/?f=a_%7B%5Cboldsymbol%7BT%7D%7D%3D%5Cfrac%7Bd%5E2s%7D%7Bdt%5E2%7D)
We know that
and ![||\boldsymbol{r}'(t)}||=2](https://tex.z-dn.net/?f=%20%7C%7C%5Cboldsymbol%7Br%7D%27%28t%29%7D%7C%7C%3D2)
so
![a_{\boldsymbol{T}}=0](https://tex.z-dn.net/?f=a_%7B%5Cboldsymbol%7BT%7D%7D%3D0)
The normal component of acceleration is given by the formula
![a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2](https://tex.z-dn.net/?f=a_%7B%5Cboldsymbol%7BN%7D%7D%3D%5Ckappa%20%28%5Cfrac%7Bds%7D%7Bdt%7D%29%5E2)
We know that
and
so
![a_{\boldsymbol{N}}=1 (2)^2=4](https://tex.z-dn.net/?f=a_%7B%5Cboldsymbol%7BN%7D%7D%3D1%20%282%29%5E2%3D4)