Answer:
Datos:
q1 = -50 μC =
q2 = +30 μC =
F = 10 N
a) x si la <em>F = 10N</em>
Aplicando la Ley de Coulomb:
x = = = 1,162m
b) x si la <em>F = 20 N</em>
x=<em> </em><em> </em>= 0,822m
c)x si la <em>F = 50 N</em>
x = = 0,520m
Answer:
If all these three charges are positive with a magnitude of each, the electric potential at the midpoint of segment would be approximately .
Explanation:
Convert the unit of the length of each side of this triangle to meters: .
Distance between the midpoint of and each of the three charges:
Let denote Coulomb's constant (.)
Electric potential due to the charge at : .
Electric potential due to the charge at : .
Electric potential due to the charge at : .
While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.
Hence, the electric field at the midpoint of due to all these three charges would be:
.
Equation C describes the sum of the vectors plotted below.
<h3>What is a vector?</h3>
A vector is a quantity or phenomena with magnitude and direction that are independent of one another. The phrase also refers to a quantity's mathematical or geometrical representation.
If no vector can be written as a linear combination of the others, a set of vectors is said to be linearly independent.
The given points from the graph is obtained as;
a = (2,1)
b = (3,-2)
Vector, OA = 2x + y
Vector, AB = x - 3 y
From the triangular lawe of the vector addition;
Hence,option C is correct.
To learn more about the vector refer to the link;
brainly.com/question/13322477
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Answer:
None, egg cells don't have chromosomes. No, sex cells do have chromosomes. Meiosis reduces chromosome number so that sex cells (eggs and sperm) have a half set of chromosomes–one homolog of each pair. This is the haploid number.
Answer:
6 days.
Explanation:
From radioactivity, The expression for half life is given as,
R/R' = 2⁽ᵃ/ᵇ)................... Equation 1
Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.
Given: R = 80 g, R' = 10 g, b = 2 days.
Substitute into equation 1
80/10 = 2⁽ᵃ/²⁾
8 = 2⁽ᵃ/²⁾
2³ = 2⁽ᵃ/²)
Equating the base and solving for a
3 = a/2
a = 2×3
a = 6 days.