An atom of an element with two valence electrons reacts with two atoms of an element with atomic number 17 to form an ionic comp
1 answer:
An atom with two valence electrons in the periodic table belongs to the alkaline earth metals or 2nd group. The element with atomic number 17 is chlorine with 7 valence electrons. Ionic compounds are formed between the atoms of elements with high electronegativity difference. More the difference, more they are likely to form ionic bond.
Thus in this case, the most likely formed ionic compound will be with alkaline earth metal with high electropositivity. The highest electropositive element in group 2 is barium thus, ionic compound formed will be
In , barium will donate its two electrons, one to each chlorine atom. The resultant compound will be ionic in which all the atom have complete octets.
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Answer:
Explanation:
The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);
where is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.
Given;
v = 0.8m/s
g =
r = ?
In order to solve for , we can simply make it the subject of formula from equation (1) as follows;
since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.
Therefore;
Answer:
71.76 m
Explanation:
We will solve this question using the work energy theorem.
The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.
ΔK.E = W
In the attached free body diagram for the question, the forces acting on the puck are given.
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J (since the puck comes to a stop)
Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J
ΔK.E = 0 - 67.6 = - 67.6 J
W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)
Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h
Workdone by the frictional force = F × d
F = μ N
μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)
N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N
F = μ N = 0.3 × 1.697 = 0.509 N
where d = distance along the incline that the puck travels.
d = h/sin 30° = 2h (from trigonometric relations)
Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h
ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)
- 67.6 = - 1.96h + 1.02h
-0.942h = - 67.6
h = 71.76 m
