The work is equal to the line integral of
over each line segment.
Parameterize the paths
- from (0, 0, 0) to (2, 0, 0) by
with
, - from (2, 0, 0) to (2, 4, 1) by
with
, - from (2, 4, 1) to (0, 4, 1) by
with
, and - from (0, 4, 1) to (0, 0, 0) by
with ![0\le t\le1](https://tex.z-dn.net/?f=0%5Cle%20t%5Cle1)
The work done by
over each segment (call them
) is
![\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7BC_1%7D%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r_1%3D%5Cint_0%5E2%5Cvec0%5Ccdot%5Cvec%5Cimath%5C%2C%5Cmathrm%20dt%3D0)
![\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7BC_2%7D%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r_2%3D%5Cint_0%5E1%28t%5E2%5C%2C%5Cvec%5Cimath%2B24t%5C%2C%5Cvec%5Cjmath%2B32t%5E2%5C%2C%5Cvec%20k%29%5Ccdot%284%5C%2C%5Cvec%5Cjmath%2B%5Cvec%20k%29%5C%2C%5Cmathrm%20dt%3D%5Cint_0%5E1%2896t%2B32t%5E2%29%5C%2C%5Cmathrm%20dt%3D%5Cfrac%7B176%7D3)
![\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7BC_3%7D%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r_3%3D%5Cint_0%5E2%28%5Cvec%5Cimath%2B%2824-12t%29%5C%2C%5Cvec%5Cjmath%2B32%5C%2C%5Cvec%20k%29%5Ccdot%28-%5Cvec%5Cimath%29%5C%2C%5Cmathrm%20dr%3D-%5Cint_0%5E2%5Cmathrm%20dt%3D-2)
![\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7BC_4%7D%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20r_4%3D%5Cint_0%5E1%28%281-t%29%5E2%5C%2C%5Cvec%5Cimath%2B2%284-4t%29%5E2%5C%2C%5Cvec%20k%29%5Ccdot%28-4%5C%2C%5Cvec%5Cjmath-%5Cvec%20k%29%5C%2C%5Cmathrm%20dt%3D-2%5Cint_0%5E1%284-4t%29%5E2%5C%2C%5Cmathrm%20dt%3D-%5Cfrac%7B32%7D3)
Then the total work done by
over the particle's path is 46.
Answer:
True
Step-by-step explanation:
If you simplify an expression you divide for example:
4 x 6 - 10 would be dividing each number by 2, 2 x 3 - 5.
I'm not a genius so hopefully this is right and helps!
Answer:
Equation of line: y=3x-18
Step-by-step explanation:
Point: (5,7) and slope=-13
y-7=-13(x-5)
y=-13x+72
Answer:
speed = 796142.197 km/hr
Step-by-step explanation:
solution
we consider here orbit around galactic center is circular so
circumference formula will be here
circumference C = 2πr ........1
here we know r = 27000
so
circumference C = 2 π × 27000
C = 169560 light year
so
we travel 169560 light year in 230000000 years
so distance will be
distance = ![\frac{169560}{230000000}](https://tex.z-dn.net/?f=%5Cfrac%7B169560%7D%7B230000000%7D)
distance = 0.00073721 light year/year
and we know 1 year = 8760 hours
so 1 light year = 9.46027 ×
km
so
sped of moving is
speed = 0.00073721 × 9.46027 ×
km/hr
speed = 796142.197 km/hr