Answer:
Since we have identical diodes we can use the equation:
And replacing we have:
Since we know that 1 mA is drawn away from the output then the real value for I would be
And for this case the value for 
would be:
And the output votage on this case would be:
And the net change in the output voltage would be:
Explanation:
For this case we have the figure attached illustrating the problem
We know that the equation for the current in a diode id given by:
![I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}](https://tex.z-dn.net/?f=%20I_D%20%3D%20I_s%20%5Be%5E%7B%5Cfrac%7BV_D%7D%7BV_T%7D%7D%20-1%5D%20%5Capprox%20I_S%20e%5E%7B%5Cfrac%7BV_D%7D%7BV_T%7D%7D)
For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode 
and each voltage is the same v for each diode, so then:
Since we have identical diodes we can use the equation:
And replacing we have:
Since we know that 1 mA is drawn away from the output then the real value for I would be
And for this case the value for would be:
And the output votage on this case would be:
And the net change in the output voltage would be:
C. Electromagnetic waves do not always need a medium to travel.
Since they are transverse waves they do not need a material medium.
Answer:
2917.4 m/s
Explanation:
From the question given above, the following data were:
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
Radius (r) of the Moon = 1737 Km
Escape velocity (v) =?
Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:
Gravitational acceleration of the earth = 9.8 m/s²
Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth
= 0.25 × 9.8 = 2.45 m/s²
Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:
1 Km = 1000 m
Therefore,
1737 Km = 1737 Km × 1000 m / 1 Km
1737 Km = 1737000 m
Thus, 1737 Km is equivalent to 1737000 m
Finally, we shall determine the escape velocity of the rocket as shown below:
Gravitational acceleration of the Moon (g) = 2.45 m/s²
Radius (r) of the moon = 1737000 m
Escape velocity (v) =?
v = √2gr
v = √(2 × 2.45 × 1737000)
v = √8511300
v = 2917.4 m/s
Thus, the escape velocity is 2917.4 m/s
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