Answer:
The magnetic field produced by an electric current is always oriented perpendicular to the direction of the current flow. And.Direction of magnetic field is governed by the 'right hand thumb rule, The right hand rule states that: to determine the direction of the magnetic force on a positive moving charge, ƒ, point the thumb of the right hand in the direction of v, the fingers in the direction of B, and a perpendicular to the palm points in the direction of Force . Similar to the situation with electric field lines, the greater the number of lines (or the closer they are together) in an area the stronger the magnetic field.
Answer:
the force applied to the smaller piston is 600 N
Explanation:
Given;
weight of the car, F = 15,000 N
radius of the lager piston, R = 0.2 m
radius of the smaller piston, r = 0.04 m
let the force applied to the smaller piston = f
The pressure applied on both piston is constant;
Therefore, the force applied to the smaller piston is 600 N
The answer to the question is (C)
Hope this helps :)
Consider that when heigt is 35 meters, the velocity is zero. The you can use the formula for free fall with initial velocity zero
Vf = √(2gh) = √(2*9.8m/s^2 * 35m) = √686m^2/s^2 = 26.2 m/s
Assumption: 35 m is the maximum height from the ground, else, if the maximum heigth from the ground is 35 m + 1 m = 36 m, the numbers are:
Vf = √(2*9.8m/s^2*36m) = 26.6 m/s
Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
