Answer:
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Explanation:
the fundamental answer is without regular supervision of definition of weights and measures,commerce exchange will be impossible and there would be no market whatsoever for anything.
Answer:
The forces experienced by the middle particle are attractive, and the net force will remain the same (0) if and only if the distances of the sides particles to the middle particle are the same.
Explanation:
In example 20.3 the forces experienced by the middle particle are repulsive because all the particles are positive, for the case in which the particles on the sides are replaced for negative charge particles the forces experienced by the middle particle are attractive. Regarding the net force, because we don't know the distances we can not give a definitive answer, what we can say is that if the distances from the middle particle to the sides particles are the same the net force is zero for both cases (remain unchanged).
Answer:
3.67 N
Explanation:
From the question given above, the following data were obtained:
Charge of 1st object (q₁) = +15.5 μC
Charge of 2nd object (q₂) = –7.25 μC
Distance apart (r) = 0.525 m
Force (F) =?
Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:
For the 1st object
1 μC = 1×10¯⁶ C
Therefore,
15.5 μC = 15.5 × 1×10¯⁶
15.5 μC = 15.5×10¯⁶ C
For the 2nd object:
1 μC = 1×10¯⁶ C
Therefore,
–7.25 μC = –7.25 × 1×10¯⁶
–7.25 μC = –7.25×10¯⁶ C
Finally, we shall determine the force. This can be obtained as follow:
Charge of 1st object (q₁) = +15.5×10¯⁶ C
Charge of 2nd object (q₂) = –7.25×10¯⁶ C
Distance apart (r) = 0.525 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²
F = 3.67 N
Therefore, the force on the object is 3.67 N
Answer:
We know that loudness is directly proportional to amplitude and SI unit of amplitude is Decibel So unit of loudness is decibel
Total distance is 28+12+34=74km=74000m.
Total time is 20+8+25 minutes=53 minutes=53×60=3180 secs.
Average speed is 74000/3180=23.27 m/s.