Find Displacement and Distance
displacement ...
north is 700+400+100 =1200m n
south=1200m
1200-1200=0
east is 300+300=600m
west is 600m
600-600=0
back at dtart. displ zero
distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m
Answer:
23.1 N/C
Explanation:
OP = 3 m , OQ = 4 m

q = - 8 nC, Q = 75 nC
Electric field at P due to the charge Q is

Electric field at P due to the charge q is

According to the diagram, tanθ = 3/4
Resolve the components of E1 along x axis and along y axis.
So, Electric field along X axis, Ex = - E1 Cos θ
Ex = - 27 x 4 / 5 = - 21.6 N/C
Electric field along y axis, Ey = E1 Sinθ - E2
Ey = 27 x 3 /5 - 8 = 8.2 N/C
The resultant electric field at P is given by

Answer:
The right answer is "The center of mass doesn't move".
Explanation:
- It generates a voltage throughout the cable while the astronaut falls on either the wire. At other ends of the spectrum or cable, the tension will be similar. As such, with both astronauts, there would be the same energy, although throughout the opposite way.
- Thus, the net force seems to be essentially negative on the machine. And therefore the mass center stays stationary.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The velocity is
in positive x -direction
The speed is 
Explanation:
From the question we are told that
The distance from the house to truck is D = 20 m
The distance traveled back to retrieve wind-blown hat is d = 15
The distance from the wind-blown hat position too the truck is k = 20 m
The total time taken is t = 75 s
Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative
Generally Justin's displacement is mathematically represented as

=> 
Generally the average velocity is mathematically represented as

=> 
=>
Generally the distance covered by Justin is mathematically represented as

=> 
=> 
Generally Justin's average speed over a 75 s period is mathematically represented as

=> 
=> 
Answer:
The electric field strength is 
Solution:
As per the question:
Area of the electrode, 
Charge, q = 50 nC = ![50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m](https://tex.z-dn.net/?f=50%5Ctimes%2010%5E%7B-%209%7D%20C%5B%2Fetx%5D%3C%2Fp%3E%3Cp%3EDistance%2C%20x%20%3D%202%20mm%20%3D%20%5Btex%5D2%5Ctimes%2010%5E%7B-%203%7D%20m)
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:

Now, the electric field strength of the electrode is:

where


