Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
Answer:
The correct solution will be "12.0 A".
Explanation:
The given values are:



By using the transformer formula, we get
⇒ 
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Answer:
1) the unit of power is A.Watt
2)100J=50W
100=50s
divide both sides by 50s
S=2s(A)
Answer:
Explanation:
The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.
The strength of electric field is defined as the force experienced by the unit positive test charge.
E = F / q
Electric field strength is a vector quantity and it is measured in newton per coulomb.
Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.
The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.
Answer:
C. 0.2 Hertz
Explanation:
The frequency of a spring is equal to the reciprocal of the period:

where
f is the frequency
T is the period
For the spring in this problem,
T = 5 s
therefore, the frequency is
