Answer:
and direction
Explanation:
because when we definite force force is the pull or push of an object and when a force is exerted on the object it works 3 tasks 1, it multiply the force , 2, it multiply the speed, 3, it changes the directions so when we see a vector quantity is a quantity that have both magnitude and direction so when it multiply it's force and speed the object will have magnitude and when it multiply it's direction the object will have a direction. thank you for reading this explanation.☺☺☺☺
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:
Explanation:
Let assume that ball's motion is frictionless and its geometric dimensions can be neglected. Besides, it is assumed that ball begins at a height of zero. The ball can be modelled by means of the Principle of Energy Conservation:
The initial speed of the ball is:
The velocity of the two balls after the collision is 0.73 m/s.
The velocity of the two balls after the collision can be calculated using the formula below.
<h3>Formula:</h3>
- mu+m'u' = V(m+m')............... Equation 1
<h3>Where:</h3>
- m = mass of the first ball
- m' = mass of the second ball
- u = initial velocity of the first ball
- u' = initial velocity of the second ball
- V = velocity of the two balls after the collision.
make V the subject of the equation
- V = (mu+m'u')/(m+m')................ Equation 2
From the question,
<h3>Given:</h3>
- m = 0.25 kg
- m' = 0.3 kg
- u = 1 m/s
- u' = 0.5 m/s
Substitute these values into equation 2
- V = [(0.25×1)+(0.3×0.5)](0.25+0.3)
- V = 0.4/0.55
- V = 0.73 m/s.
Hence, the velocity of the two balls after the collision is 0.73 m/s
Learn more about collision here: brainly.com/question/7694106