If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.
Reasoning:
It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.
Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.
Thus the electric force applied on the charge in my hand due to each other is,
Here k is the Coulomb constant, and r is the distance between the charges.
It is also known that the force on a positive charge due to another positive charge is acted outwards.
Thus, the force on the charge due to the charge on the east is,
And the force on the charge due to the charge on the north is,
As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 
from the north or from the north direction.
Thus the net force is acting in the north-east direction.
Learn more about the direction of the force here,
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Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
where m is the slope
Comparing equation (1) and (2)
a = m
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s
The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
![[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%2822%20-%2011.78%29%20%2A%20%287.30%29%5D%20%20%2B%20%5B%2811.78%20-%200%29%20%2A%20%287.30%29%5D)
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
No image will be observed.
Explanation:
Images that are created by mirrors are virtual images. This virtual image can only be seen by an observer. In this case, an infinite number of images or no image will be created here as both will be reflecting their own images. Light will continuously bounce back and forth reflecting the same image.
The bending of a wave as it crosses a boundary between 2 mediums at an angle is called refraction.
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>
