Answer: electronegativity
Explanation:
Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.
The size of an atom decreases as we move across the period because the electrons get added to the same shell and the nuclear charge keeps on increasing. Thus the electrons get more tightly held by the nucleus.
As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases and thus the electronegativity increases.
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):

Now At xp we have:


Which is a second order equation, using the quadratic formula to solve for xp would give us:

or

Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that
BOTH are correct. This is simply explained by considring the following.
Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.
Answer:
a) k = 891.82 N/m
b) e = 0.0143 m = 1.43 cm
c) W = 5.02 J
Explanation:
Step 1: Data given
Mass = 2.60 kg
the spring stretches 2.86 cm = 0.0286
Step 2: What is the force constant of the spring?
Force constant, k = force applied / extension produced
k = (2.60kg * 9.81N/kg) / 0.0326 m
k = 891.82 N/m
b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it
Extension = F/k = (1.30 kg * 9.81) / 891.82 = 0.0143 m = 1.43 cm
Half the mass means half the extension
c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?
W = average force used * distance
W = 1/2 * k*e * e = 1/2 k*e²
W = 1/2 * 891.82 * (0.075)² = W = 5.02 J
very interesting and amazing question u had asked but i get it...
Answer
W = Force × Displacement
- but in above case force is exerted on the men as the weight of the child × gravitational pull of earth (9.8 m/s^2) but the displacement or movement is ZERO
- there for W= F×0 =0
- so even after carrying child for 2 hours work done by the men will be zero
<u>More</u><u> </u><u>examples</u>
- a person standing straight and lifting weight
- a person carry some weight and walk in a circular path and came back to its original position.........in this case also Displacement will be zero
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hope it helped u buddy