Answer:
it is attracted to negative pole
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
Work done = Potential Energy = (mg)h
mg = weight of the ball
= 1000 N * 2.5 m = 2500 Joules.
The work done during the 5 second is 2500 Joules.
Answer:
the body has linear acceleration, but cannot rotate
Explanation:
Let's analyze the system
If the torque is zero, the two forces are the same magnitude, but applied to each side of the body in such a way that the torque cancels the punch of the other. Therefore the body cannot turn
The two forces go in the same direction so the object can have linear acceleration
The object is at rest because it has a force in the same direction, but in the opposite direction.
therefore the correct answer is:
the body has linear acceleration, but cannot rotate
As per the question the distance of venus from sun is given as 0.723 AU
We have been asked to calculate the time period of the planet venus.
As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

⇒
where is k is the proportionality constant
We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days
Hence
The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun
Hence 
The distance of venus from sun is 0.723 AU i.e
From keplers law we know that-
⇒
Putting the values mentioned above we get-

⇒
⇒
Hence the time period of venus is 224.388352752710 days