To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.
Where,
Q = Heat
U = Internal Energy
By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore
Work is done ON the system
Heat is extracted FROM the system
Therefore the value of the Work done on the system is -158.0J
Answer:
The conduction path or simply the wires connected between different components in a circuit.
Explanation:
The wire makes up the path for the electricity to flow and most of the electricity flows through this. It is like a road connecting two house or buildings in a town and the traffic of vehicles is the electricity (current).
It's called buoyancy. It is the tendency of an object to float
<span>C. It is the difference in electrical potential energy between two places in an electric field.</span>
<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
