Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
Star 1 - 4 hours right ascension
Star 2 - 3 hours right ascension
Subtracting hours right ascension
4 hours right ascension - 3 hours right ascension = 1 hours right ascension.
Thus,
star 1 will rise 1 hour before star 2
Its B: reduce the amount of energy needed to do the work by putting the work onto something else
Answer:
<em>OPTRIMUM</em><em> </em><em>PRIDE</em><em> </em><em>URGH</em><em> </em><em>URGH</em><em> </em><em>URGH</em><em> </em>
Explanation:
AHHAAHAHAHAHA
To solve this problem we will apply the concepts related to gravitational potential energy.
This can be defined as the product between mass, gravity and body height.
Mathematically it can be expressed as
Therefore the change in the internal energy of the system is 255.78
