Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
can you have all of them as possibilities ???
Answer options:
A) Some connections are lost because critical stages of development were missed.
B) Some connections are lost so others can be strengthened.
C) Some connections are lost because of experimentation with drugs or alcohol.
D) Some connections are lost so that children become less dependent and are less tied to their parents.
Answer:
B) Some connections are lost so others can be strengthened.
Explanation:
Synaptic pruning refers to a natural process that occurs in the brain between childhood and adulthood. Extra synaptic connections are eliminated in order to increase the efficiency of neuronal communications. It is thought to be the brain’s way of removing connections that are no longer needed.
The answer would be C. It will decrease with descent. Hope this helps!
Given E=2050 N/C
charge of an electron = 1.6×10^-19 C
USING E = F/Q , electric field = force ÷ charge
2050 = F ÷ (1.6×10^-19)....
make F subject of formula and get the answer
