Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>
Answer:
-13.18°C
Explanation:
To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.
Its definition is given by the function

Where,
Q = The amount of heat transferred
t = time
k = Thermal conductivity constant
A = Cross-sectional area
The difference in temperature between one side of the material and the other
d= thickness of the material
The problem says that there is a loss of heat twice that of the initial state, that is

Replacing,




Solvinf for
,

Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C
Answer:
Energy needed = 1100 kJ
Explanation:
Energy needed = Change in kinetic energy
Initial velocity = 15 m/s
Mass, m = 1600 kg

Final velocity = 40 m/s

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J
Energy needed = 1100 kJ
Answer:
b. 9.5°C
Explanation:
= Mass of ice = 50 g
= Initial temperature of water and Aluminum = 30°C
= Latent heat of fusion = 
= Mass of water = 200 g
= Specific heat of water = 4186 J/kg⋅°C
= Mass of Aluminum = 80 g
= Specific heat of Aluminum = 900 J/kg⋅°C
The equation of the system's heat exchange is given by

The final equilibrium temperature is 9.50022°C
Answer:
the inductive reactance of the coil is 1335.35 Ω
Explanation:
Given;
inductance of the coil, L = 250 mH = 0.25 H
effective current through the coil, I = 5 mA
frequency of the coil, f = 850 Hz
The inductive reactance of the coil is calculated as;

Therefore, the inductive reactance of the coil is 1335.35 Ω