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>
Explanation:
Newton's second law shows that there is a direct relation ship between force and acceleration . the grater force that is applied on a object of given mass the more the accelerate. for example doubling the force in the object doubles it's acceleration.
Answer:
The answer will be B
Explanation:
When the image is erect and magnified it means that the magnification is positive and greater than 1.
The mirror formula is 1u+1v=1f, where,
u is the distance of the object from the mirror,
v is the distance of the image from the mirror, and,
f is the focal length of the mirror.
⇒1v=1f−1u=u−fuf.
⇒v=ufu−f.
The magnification, M=−vu.
⇒M=−fu−f=ff−u
Also, the magnification is M=height of the imageheight of the object.
As per the sign convention, in a concave mirror, u,v and f are all negative, the height or an erect image is positive and the height of an inverted image is negative.
Hence, for an erect image, it is necessary to have M>0.
⇒M=ff−u>0.
⇒f−u<0 since f<0.
⇒f<u.
⇒|u|<|f| since f<0.
⇒ The distance of the object from the mirror is lesser than the focal length.
Further, |f|>|f−u| since both f and u both have the same sign.
⇒ff−u>1.
⇒ The image is magnified i.e. the height of the image is greater than the height of the object.
In your question where the ask is to calculate the charge that the small sphere carries which is the mass of it is 441g moving at an acceleration of 13m/s^2 nad having and electric field of 5N/C. So the formula in getting the charge is mutliply the mass and the quotients of Acceleration and the Electric Field so the answer is 1,146.6