electric field lines are graphical presentation of electric field intensity
It is the graphical way to represent the electric field variation
If we draw the tangent to electric field line then it will give the direction of net electric field at that point
So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties
here we will always follow these basic properties of field lines
now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined
As we have two directions of tangents at that point
So here the incorrect presentation is the intersection of two field lines which is not possible
Answer:
The horizontal component of the velocity is 188 m/s
The vertical component of the velocity is 50 m/s.
Explanation:
Hi there!
Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:
We can find vx using the following trigonometric rule of a right triangle:
cos α = adjacent / hypotenuse
cos 15° = vx / 195 m/s
195 m/s · cos 15° = vx
vx = 188 m/s
The horizontal component of the velocity is 188 m/s
To calculate the y-component we will use the following trigonometric rule:
sin α = opposite / hypotenuse
sin 15° = vy / 195 m/s
195 m/s · sin 15° = vy
vy = 50 m/s
The vertical component of the velocity is 50 m/s.
Answer:39.88 rad/s
Explanation:
Given
mass of cylinder m_1=18 kg
radius R=1.7 m
angular speed 
mass of
dropped at r=0.3 m from center
let
be the final angular velocity of cylinder
Conserving Angular momentum





Answer:
Explanation:
Height covered = 12m
time to fall by 12 m
s = 1/2 gt²
12 = 1/2 g t²
t = 1.565 s
Horizontal distance of throw
= 8.5 x 1.565
= 13.3 m
This distance is to be covered by dog during the time ball falls ie 1.565 s
Speed of dog required = 13.3 / 1.565
= 8.5 m /s
b ) dog will catch the ball at a distance of 13.3 m .