The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
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Answer:
Explanation:
5.non contact force
6.balanced force
7.unbalanced force
8.net force
9.zero
10.is the difference between two forces
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Explanation:
Given data:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Distance d between the plates = 1 mm = 1×10⁻³m
Voltage of the battery is emf = 100 V
Resistance = 1025 ohm
Solution:
In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

Taking natural log on both sides,

(1)
Now we can calculate the capacitance by using the area of the plates.
C = ε₀A/d
= 
= 18×10⁻¹²F
Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
Answer:
Q = 1.35*10⁻¹¹ C.
Explanation:
By definition, the capacitance of a capacitor, is the charge on one of the plates, divided by the potential difference between them, as follows:

At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:
C = ε₀*A / d
Replacing by the values of A, and d, and taking into account that
ε₀ = 8.85*10⁻¹² F/m,
we get the value of the capacitance as follows:
C = 8.97*10⁻¹² F
As the voltage of an AA battery is 1.5 V, and is all applied to the capacitor, we can conclude that the charge on one of the plates is as follows:
Q = C* V = 8.97*10⁻¹² F* 1.5 V = 1.35*10⁻¹¹ C