The ultimate source of electromagnetic waves are the radio sets. The answer is letter C. <span>One example of an
electromagnetic radiation is the visible light. And visible light can be radio
waves, infrared light and X - rays. The rays of the sun are a form of visible
light. It has an electromagnetic radiation of UV (ultra violet) rays. That is
why the radiation at day is greater than at day due to sun’s rays.</span>
Answer:
5.4×10⁶J
Explanation:
1 cal = 4.184 J
1.3×10⁶ cal × (4.184 J/cal) = 5.4×10⁶J
Answer:
C
Explanation:
The change in momentum of x has to be the opposite of the change in momentum of Y because the momentum is just transferred from one to another. But I'm still trying to figure it out how to calculate.
There are several information's of immense importance already given in the question. Based on the given information's the answer to the question can easily be determined.
Distance covered by the bicycle = 5000 meter
Time taken by the bicycle to reach the distance = 500 second.
Velocity of the bicycle = Distance / Time taken
= 5000/500 meter/second
= 50 meter/second
So the velocity of the bicycle is 50 meter per second. I hope the procedure is clear enough for you to understand. In future you can always use this procedure for solving similar problems.
Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
