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sp2606 [1]
3 years ago
9

Two charges, q1 and q2, are separated by a certain distance r. if the magnitudes of the charges are halved and their separation

is halved, then what happens to the electrical force between these two charges?
Physics
1 answer:
vazorg [7]3 years ago
6 0

The electrical force between these two charges remains the same. In coulomb’s law, it states that the magnitude of two charges (product of two charges) is inversely proportional to the square of the distance. Since both the magnitude and the distance are halved, therefore, the change in both quantities will have no effect in the value of electrical force.

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A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p
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b) 833.3 s

c) Because of proportions

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t_{total}=t_{up}+t_{down}

Time upriver is calculated with the net speed of student and 0.500 km:

t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s

(Becareful with units 0.5 km= 500m) Similarly of downriver:

t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s

So the sum is:

t_{total}=1111.0s

b) Still water does not affect student speed, so total time would be simply:

t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.  

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