Complete question is;
A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.
Answer:
Power ≈ 600,000 W
Explanation:
We are given;
Frequency; f = 2400 Hz
height of the oven cavity; h = 25 cm = 0.25 m
base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²
total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J
We want to find the power output and we know that formula for power is;
P = workdone/time taken
Formula for time here is;
t = h/c
Where c is speed of light = 3 × 10^(8) m/s
Thus;
t = 0.25/(3 × 10^(8))
t = 8.333 × 10^(-10) s
Thus;
Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))
Power ≈ 600,000 W
Impulse in physics is the integral of force, F, with respect with time, t. This value is a vector quantity since force is a vector quantity as well. It can be calculated from the product of force and time. We do as follows:
Impulse = Ft
= m(a)(t)
= m(v/t)t
= 0.046 (42/0.0005) (0.0005)
= 1.932 N-s
The back-and-forth movement of electrons is called alternating current. Electrons go back and forth, the direction of their path alternates from one direction to another.
the movement of electrons in one direction is called direct current. The electrons move in a direct, single path without changing directions.
Heat engines are less than 100% efficient because absolute zero cannot be reached
Answer:
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
, r is radius
Let I is the displacement current. It is given by :
Here, 
is rate of increasing potential difference
So
So, the value of displacement current is 
.
