Answer:
Pascal's law (also Pascal's principle or the principle of transmission of fluid-pressure) is a principle in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.
Explanation:
2) C would need the least effort, because the longer the effort distance, the least the effort applied.
To find the tangent plane to the surface f(x,y,z)=0 at a point (X,Y,Z) we use the following method:
<span>Calculate grad f = (f_x, f_y, f_z). The normal vector to the surface at the point (X,Y,Z) is grad f(X,Y,Z). The equation of a plane with normal vector n which passes through the point p is (r-p).n=0, where r=(x,y,z) is the position vector. So the equation of the tangent plane to the surface through the point (X,Y,Z) is ((x,y,z)-(X,Y,Z)).grad f(X,Y,Z)=0. </span>
<span>Now in your case we have f(x,y,z)=y-x^2-z^2, so grad f=(-2x,1,-2z), and the equation of the tangent plane at the point (X,Y,Z) is </span>
<span>((x,y,z)-(X,Y,Z)).(-2X,1,-2Z)=0, </span>
<span>that is </span>
<span>-2X(x-X)+1(y-Y)-2Z(z-Z)=0, </span>
<span>i.e. </span>
<span>-2Xx+y-2Zz = -2X^2+Y-2Z^2. (1) </span>
<span>Now compare this equation with the plane </span>
<span>x + 2y + 3z = 1. (2) </span>
<span>The two planes a_1x+b_1y+c_1z=d_1, a_2x+b_2y+c_2z=d_2 are parallel when (a_1,b_1,c_1) is a multiple of (a_2,b_2,c_2). So the two planes (1),(2) are parallel when (-2X,1,-2Z) is a multiple of (1,2,3), and we have </span>
<span>(-2X,1,-2Z)=1/2(1,2,3) </span>
<span>for X=-1/4 and Z=-3/4. On the paraboloid the corresponding y coordinate is Y=X^2+Z^2=1^4+9^4=5/2. </span>
<span>So the tangent plane to the given paraboloid at the point (-1/4,5/2,-3/4) is parallel to the given plane.</span>
Answer:
17.2 seconds
Explanation:
Given:
v₀ = 0 m/s
a₁ = 10.0 m/s²
t₁ = 3.0 s
a₂ = 16 m/s²
t₂ = 5.0 s
a₃ = -12 m/s²
v₃ = 0 m/s
Find: t
First, find v₁:
v₁ = a₁t₁ + v₀
v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)
v₁ = 30 m/s
Next, find v₂:
v₂ = a₂t₂ + v₁
v₂ = (16 m/s²) (5.0 s) + (30 m/s)
v₂ = 110 m/s
Finally, find t₃:
v₃ = a₃t₃ + v₂
(0 m/s) = (-12 m/s²) t₃ + (110 m/s)
t₃ = 9.2 s
The total time is:
t = t₁ + t₂ + t₃
t = 3.0 s + 5.0 s + 9.2 s
t = 17.2 s
Round as needed.