
 Actually Welcome to the Concept of the kinematics of a body.
Since, we know that Velocity = Distance / time
hence, V = 20/5 = 4 m/s 
hence the velocity of the RC car is 4 m/s westwards direction.
 
        
             
        
        
        
Answer:

Explanation:
For answer this we will use the law of the conservation of the angular momentum.

so:

where  is the moment of inertia of the merry-go-round,
 is the moment of inertia of the merry-go-round,  is the initial angular velocity of the merry-go-round,
 is the initial angular velocity of the merry-go-round,  is the moment of inertia of the merry-go-round and the child together and
 is the moment of inertia of the merry-go-round and the child together and  is the final angular velocity.
 is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where  is the mass and R is the radio of the merry-go-round
 is the mass and R is the radio of the merry-go-round 
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2 rad/s
 rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:

 

Finally we replace all the data:

Solving for  :
:

 
        
             
        
        
        
Answer:
please the answer below
Explanation:
(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

k=8.89*10^9 
For both cases we have

(b) by replacing this values of r in the expression for V we obtain

hope this helps!!
 
        
                    
             
        
        
        
Answer:
Open- closed
Explanation:
It has Open-closed configuration of its end
 
        
             
        
        
        
Answer:
Explanation:
Its definitely an Attractive force since the two charges are Unlike. 
From Coulombs Law 
F=kq1q2/R²
Given 
K=9x10^9
R=1m
q1=2C 
q2=-1C
F=(9x10^9 x 2 x -1)/1²
F= - 1.8x10^10N. (Attractive).