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 
Answer:
Two charged objects have a repulsive force of 0.080 N. If the charge of both of the objects is doubled, then what is the new force? Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of both objects is doubled, then the force will become four times greater.
Explanation:
hope this helps
<span>An alpine glacier can change the topography of a mountainous area through Glacial Erosion and Glacial Deposition. Glaciers are agents of erosion, it can pick up and carry large rocks and sediments. In the process, a deep cavity or hole can form when the glacier plucks a big rock from where it passed. Glaciers have shaped many Mountain Ranges and have created distinct landforms by its erosion process. In Glacial Deposition, as glaciers melt, it deposits all that it carried and a landform is developed.</span>
Answer:

Explanation:
information we know:
Total force: 
Weight: 
distance: 
vertical component of the force: 
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In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).
horizontal component: 
vertical component: 
but from the given information we know that 
so, equation these two
and 

and we know the force
, thus:

now we clear for 

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:


whith this horizontal component we calculate the work to move the crate a distance of 4 m:

the work done is W=173.48J
Answer:
Explanation:
The sandpaper block did not move because the forces of friction and gravity were balanced.