The intensity of the electric field is 30,000 N/C
Explanation:
The strength of the electric field produced by a single-point charge is given by the equation
 
where:
 is the Coulomb's constant
 is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
In this problem, we have:
 is the magnitude of the charge
 is the magnitude of the charge
r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity
Substituting, we find:

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Answer:
22Volts
Explanation:
The pd at the terminal is known as the emf
Since there are Ten 2.2V cells
Terminal voltage = number of cells * pd of one cell
Terminal voltage = 10 * 2.2
Terminal voltage = 22V
Hence the pd at the battery terminals is 22Volts
 
        
             
        
        
        
The answer should be B. A half step
 
        
                    
             
        
        
        
Explanation:
It is given that,
A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.
In 2 seconds, distance covered by the mass is 12 cm.
In 1 seconds, distance covered by the mass is 6 cm
So, in 16 seconds, distance covered by the mass is 96 cm
So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.