One can solve the problem by using the law of conservation of momentum. The total momentum prior to the collision must be equivalent to the total momentum after the collision, so we have:
m1v1 + m2v2 = m1v1 + m2v2
Here, m1 is 0.4 Kg that is the mass of the ball, u1 is 18 m/s that is the initial velocity of the ball, m2 is 0.2 Kg that is the mass of the bottle, and u2 is 0 that is the initial velocity of the bottle.
v1 is the final velocity of the ball, which is to be determined, and v2 is 25 m/s that is the final velocity of the bottle.
Substituting and rearranging the equation, one can find the final velocity of the ball:
v1 = m1u1 - m2v2 / m1 = (0.4 kg) (18 m/s) - (0.2 Kg) (25 m/s) / 0.4 Kg = 5.5 m/s.
Identifies the number of protons a single atom of the element contains.
The atomic number helps people identify elements according to the number of protons one atom of the element has. It essentially defines the element
A nuclear reaction changes the number of protons and or neutrons in an atom.
<h3>What do nuclear reactions change in the atom?</h3>
A nuclear reaction is a type of reaction that results in the change of the nucleus of an atom. We know that a nucleus of an atom is made up of protons and neutrons.
So we can conclude that a nuclear reaction changes the number of protons and or neutrons in an atom.
Learn more about reaction here: brainly.com/question/26018275
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4 adolescence occurs between childhood and adulthood
If we abbreviate the formula for nicotine as Nic, then the equations for two different equilibria of Nic in water are
Nic + H2O ---> NicH+ + OH-
NicH+ + H2O ---> NicH2 2+ + OH-
We can write the Kb1 expression for the first equation as
Kb1 = 1.0×10^-6 = [NicH+][OH-] / [Nic]
1.0×10^-6 = x^2 / 1.85×10^-3 - x
Approximating that x is negligible compared to 1.85×10^-3 simplifies the equation to
1.0×10^-6 = x^2 / 1.85×10^-3
x = 0.0000430
x = [OH-] = 4.30×10^-5 M
From the Kb2 expression
Kb2 = 1.3×10-11 = [NicH2 2+][OH-] / [NicH+]
1.1×10^-10 = x^2 / 4.30×10^-5 - x
Approximating that x is negligible compared to 4.30×10^-5 simplifies the equation to
1.1×10^-10 = x^2 / 4.30×10^-5
x = [OH-] = 6.88×10^-8
The concentration [OH-] can be computed as
[OH-] = 4.30×10^-5 M + 6.88×10^-8 M = 4.30×10^-5 M
This shows that the second equilibrium has a negligible effect on the pH.
We can now calculate for pH:
pOH = -log [OH-] = -log (4.30×10^-5 M) = 4.37
pH = 14 - pOH = 14 - 4.37 = 9.63
