Answer:
a.Attractive

Explanation:
When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.
Here, there is a proton and electron which are different particles hence, they will attract each other.
= Charge of electron and proton = 
r = Distance between them = 997 nm
k = Coulomb constant = 
Force is given by

The force of attraction between the particles will be 
If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.
Given,
Tension force in the rope is (T) = 160 N
Displacement of the skier (S) = 270 m
The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.
Therefore, work done by the rope on the skier is,

⇒
Hence work done by the rope is - 43200 J.
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Answer:
potential energy increases.
Explanation:
The potential energy between the two charged particles is given by
U = k Q q / r
If they are very far apart then r tends to infinity and the potential energy is zero.
If they come closer then the potential energy between the two charged particles increases.
Thus, the potential energy increases.
Answer:
Mass remains constant but weight reduces
Explanation:
Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.
Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.
Therefore, for this case, since g decreases, the weight decreases but mass remains constant.
Answer:
E = 31.329 N/C.
Explanation:
The differential electric field
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where
is the radius of curvature.
Now
,
where
is the charge per unit length, and it has the value

Thus, the electric field at the center of the curvature of the arc is:


Now, we find
and
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

and this is
radians.
Therefore,

evaluating the integral, and putting in the numerical values we get:

