Answer:
c. Ф
increases and E decreases
Explanation:
From Gauss's Law
Ф
= Q / ε₀ _____(1)
The electric flux Ф
is the surface integral of electric field E
Ф
_____(2)
|E| = Q/ 4πε₀r² _____(3)
Where,
Φ
is the electric flux through a closed surface S enclosing any volume V
E is the electric field
dA is a vector representing an infinitesimal element of area of the surface
Q is the total charge enclosed within V, and
ε₀ is the electric constant
Considering the relationship between equation 1, 2 and 3, If the charge, Q, is doubled and the radius, r, of the surface is also doubled; the electric flux Φ
increases and electric field E decreases. From Gauss law in equation 1, the electric flux Φ
is directly proportional to the charge, Q, with ε₀ being a constant. An increase in charge will cause an increase in electric flux. Also, the electric field, will decrease because of the inverse relationship between radius and electric field.
Answer: 1.0
Explanation:
not sure why but that is the correct answer
Answer:
200 K
Explanation:
0 °C = 273 K
-73°C = 273 K - 73 K = 200 K
It is boiling point for celsius
A. At the top. At this point, the child is highest, and they are not moving. Their potential energy is much greater.