The two neutral atoms A and B have the same number of electrons and atomic number 11. So, the two elements are said to be same.
The electronic configuration of the element is the arrangement of the electrons in the atom of the element in energy levels, orbitals around the nucleus.
The electrons in the atoms of the element with lowest energy are written first before those with higher energy levels. Thus, the electronic configuration shows the electrons in the atoms of the element arranged in order of increasing energies.
The electronic configuration of atoms are given as
A = 1s² 2s² 2p⁶ 3s¹
B = 1s² 2s² 2p⁶ 5s¹
The number of electrons in both the elements is 11. Therefore, their atomic number is also the same i.e, 11. So, both the elements are the same.
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Answer:Newton's 2nd law
Explanation:Newton's 2nd law also states that the rate at which an object changes speed is proportional to the force that is exerted. Engines provide thrust and accelerate a plane forward along the runway. If the engines supply a small force, only a small acceleration will result.
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>
The displacement of the car after the given final velocity is 31.59 m.
The given parameters;
- Initial velocity of the car, u = 24.5 m/s east
- Final velocity of the car, v = 18.3 m/s east
- Acceleration of the car, a = 4.2 m/s²
The displacement of the car after the given final velocity is obtained by applying the third kinematic equation as shown below;
v² = u² + 2as
Thus, the displacement of the car after the given final velocity is 31.59 m.
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