The potential difference between two points is 14.9 V when a 0.0400 C charge moves between these points by how much does its pot
1 answer:
Answer:
The potential energy change is 0.596 J.
Explanation:
Given that,
Potential difference =14.9 V
Charge q =0.0400 C
We need to calculate the change potential energy
The potential energy change is the product of the charge and potential difference.
Using formula of change potential energy
Hence, The potential energy change is 0.596 J.
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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.
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.
Answer:
The power dissipated in either one of the parallel resistors is 2 V
Explanation:
Given;
two parallel resistors, R₁ and R₂ = 2 ohms
The total resistance of the Two resistors of 2 ohms connected in parallel is;
when connected to another resistor of 1 ohm in series, the total resistance becomes;
Rt = R₁ + R₂
Rt = 1 + 1 = 2 ohms
Current in the circuit, I = voltage / total resistance
= 2 /2 = 1 A
the overall circuit has been resolved to series connection, and current flow in series circuit is constant.
Power = I²R
Thus, power dissipated in either one of the parallel 2 ohms resistors is;
Power = I²R = (1)² x 2 = 2 V