Answer:
6787.5 V
Explanation:
From the question,
P = IV..................... Equation 1
Where P = Power, I = rms current, V = rms voltage.
make V the subject of the equation
V = P/I................. Equation 2
Given: P = 1500 W, I = 6.4/√2 = 4.525 A
Substitute these values into equation 2
V = 1500(4.525)
V = 6787.5 V
Hence the rms voltage = 6787.5 V
The correct answer should be bounced off since the light ray hit the surface and reflected towards a new location. It therefore bounced off.
-- pick a planet from the table
-- take it's mass and radius from the table, and plug them into the big ugly formula above the table
-- do the arithmetic with your pencil or your calculator. The answer is the acceleration of gravity on the planet you picked. Write it down so you don't lose it.
-- do the same for the other 3 planets in the table
780 seconds, or 13 minutes.
In the future, please use proper capitalization. There's a significant difference in the meaning between mV and MV. One of them indicated millivolts while the other indicates megavolts. For this problem, I'll make the following assumptions about the values presented. They are:
Total energy = 1.4x10^11 Joules (J)
Current per flash = 30 Columbs (C)
Potential difference = 30 Mega Volts (MV)
First, let's determine the power discharged by each bolt. That would be the current multiplied by the voltage, so
30 C * 30x10^6 V = 9x10^8 CV = 9x10^8 J
Now that we know how many joules are dissipated per flash, let's determine how flashes are needed.
1.4x10^11 / 9x10^8 = 1.56E+02 = 156
Since each flash takes 5 seconds, that means that it will take about 5 * 156 = 780 seconds which is about 780/60 = 13 minutes.
Answer:
Zero
Explanation:
The work done by a force on an object is given by:
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and the displacement of the object
In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so 
and 
. Therefore, the work done by gravity is also zero.
