There is a few things. The things that turn it red is acids.
Properties of acids include:
*Tasting sour
*Conduct electricity
*Corrosive
A few household acids would be lemons, vinegar, salt, and some sodas.
Answer:
elastic partial width is 2.49 eV
Explanation:
given data
ER E = 250 eV
spin J = 0
cross-section magnitude σ = 1300 barns
peak P = 20ev
to find out
elastic partial width W
solution
we know here that
σ = λ²× W / ( E × π × P ) ...................1
put here all value
σ = (0.286)² × W ×
/ ( 250 × π × 20 )
1300 ×
= (0.286)² × W ×
/ ( 250 × π × 20 )
solve it and we get W
W = 249.56 ×
so elastic partial width is 2.49 eV
Answer:
FC vector representation

Magnitude of FC

Vector direction FC
degrees: angle that forms FC with the horizontal
Explanation:
Conceptual analysis
Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.
The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.
degrees
To calculate the magnitudes of the forces we apply Coulomb's law:
Equation (1): Magnitude of the electric force of the charge qA over the charge qC
Equation (2)
: Magnitude of the electric force of the charge qB over the charge qC
Known data





Problem development
In the equations (1) and (2) to calculate FAC Y FBC:


Components of the FBC force at x and y:


Components of the resulting force acting on qC:


FC vector representation

Magnitude of FC

Vector direction FC
degrees: angle that forms FC with the horizontal