Explanation:
False
 electron and proton attract each other
 
        
             
        
        
        
Answer:
The time is 1.8s
Explanation:
The ball droped, will freely fall under gravity.
Hence we use free fall formula to calculate the time by the ball to hit the ground

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.
From the question,
h=16m
Also, let take 

By substitution we obtain,


Diving through by 9.8


square root both sides, we obtain


 
        
             
        
        
        
Answer:
 You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
         k  = 50.2
        H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
        A  =  ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
        A  = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
        A  =  = 6.77 ×
 = 6.77 × m²
 m²
Now Area of cylinder is :
      A =   d²
 d²
solving for d:
     d =  
     d  = 9.28 cm
 
        
             
        
        
        
Explanation:
The given data is as follows.
        Angular velocity ( ) = 2.23 rps
) = 2.23 rps
      Distance from the center (R) = 0.379 m
First, we will convert revolutions per second into radian per second as follows.
              = 2.23 revolutions per second
              =  
              = 14.01 rad/s
Now, tangential speed will be calculated as follows.
   Tangential speed, v =  
                                = 0.379 x 14.01
                                = 5.31 m/s
Thus, we can conclude that the tack's tangential speed is 5.31 m/s.
 
        
             
        
        
        
The answer is approximately 2.998e+8