The correct answer is the First choice
Answer:
E_aprox = 1.003 E_real
Explanation:
In this exercise we are given the expression for the electric field of a dipole in the axis direction of the dipole
E_real = k 2q d / √(z² + d²)³
I think your equation has some errors.
In this case they indicate that d is the separation of the charges of the dipole
in the case of z »d this equations approximates
E_aprox = k 2q d/ z³
calculate the value for the two cases
E_real = k2q d / √[ ((23d)² + d²)³]
E_real = k2q d / d³ 12201
E_real = k2q 1/12201 d²
E_aprox = k2q d / (23.00d)³
E_aprox = k2q 1/12167 d²
the error between these quantities is
E_aprox / E_real = 12201 d² / 12167 d²
E_aprox / E_real = 1.003
E_aprox = 1.003 E_real
Answer:
Here gravitational force is G or gravity constant.
Answer:
a) 20s
b) 500m
Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.
To find time, we apply the UARM formula:
v final = (a x t) + v initial
Replacing the values gives us:
0 = (-10 x t) + 100
-100 = -10t
t = 10s
It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur
So 10s going up and another 10s going down:
10x2 = 20s
b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:
Δy = (1/2)(a)(t^2) + (v initial)(t)
Replacing the values gives us:
Δy = (1/2)(-10)(10^2) + (100)(10)
= (-5)(100) + 1000
= -500 + 1000
= 500 m
Hope this helps, brainliest would be appreciated :)
