Basic solutions are hydroxides therefore the answer is A ca(OH)2
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
C is the diffraction angle.... step by step explanation= I think it’s that I might be wrong lol
A force vector F1 points due
east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The
resultant of the two vectors has a magnitude of 400 newtons and points along
the due east/west line. Find the magnitude and direction of F2. Note that there
are two answers.
<span>The given values are
F1 = 200 N</span>
F2 =?
Total = 400 N
Solution:
F1 + F2 = T
200 N + F2 = 400N
F2 = 400 - 200
F2 = 200
N
