Answer:
Now e is due to the ring at a
So
We say
1/4πEo(ea/ a²+a²)^3/2
= 1/4πEo ea/2√2a³
So here E is faced towards the ring
Next is E due to a point at the centre
So
E² = 1/4πEo ( e/a²)
Finally we get the total
Et= E²-E
= e/4πEo(2√2-1/2√2)
So the direction here is away from the ring
Answer:
C
Explanation:
I got it right on the test !!
Answer:
c > √(2ab)
Explanation:
In this exercise we are asked to find the condition for c in such a way that the results have been real
The given equation is
½ a t² - c t + b = 0
we can see that this is a quadratic equation whose solution is
t = [c ±√(c² - 4 (½ a) b)] / 2
for the results to be real, the square root must be real, so the radicand must be greater than zero
c² -2a b > 0
c > √(2ab)
The ball because the Kinetic Energy transfers from the bat to the ball, increasing the movement and acceleration of the ball because of the Kinetic Energy transferred from the origin force (The bat)
Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment 
Mass of the heavier fragment 
Speed of lighter fragment = 0.893c
We need to calculate the speed of the heavier fragment
Let v is the speed of the second fragment after decay
Using conservation of relativistic momentum
Hence, The speed of the heavier fragment is 0.335c.
