Answer: P = I2R = 0.032 x 1000 =0.9 W
Explanation: The power will be the product of the square of the current and
the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.
Answer:
total width bandwidth = 8kHz
Explanation:
given data
transmitter operating = 3.9 MHz
frequencies up to = 4 kHz
solution
we get here upper side frequencies that is
upper side frequencies = 3.9 ×
+ 4 × 10³
upper side frequencies = 3.904 MHz
and
now we get lower side frequencies that is
lower side frequencies = 3.9 ×
- 4 × 10³
lower side frequencies = 3.896 MHz
and now we get total width bandwidth
total width bandwidth = upper side frequencies - lower side frequencies
total width bandwidth = 8kHz
Answer:
The rate of energy absorbed per unit time is 3500W.
Explanation:
From the question, we were given the following parameters;
Plane, opaque, gray, diffuse surface
â = 0.7
Surface area, A = 0.5m²
Incoming radiant energy, G = 10000w/m²
T = 500°C
Rate of energy absorbed is âAG;
âAG = 0.7 × 0.5 × 10000
âAG = 3500W.
The energy absorbed is measured in watts and denoted by the symbol W.
Answer:
A range from 5,000 miles to 7,500 miles, on average. Some recommended intervals may be shorter or longer.