Answer:to revise or edit
anything that can be made in the non draft one
Explanation:
Answer:
agricultural production of food
Explanation:
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
Answer:
pretty sure its motor 2, with body 1
Explanation:
<span>assuming the pitch is 100yards long, the player runs 100yards to the other goal then a further 50 yards back to the 50-yard line. So he/she runs 150yards in 18s
150/18 = 8.33yards per second average speed.
Initial velocity = 0, average velocity =8.33
Vav = (Vinitial+Vfinal)/2
Vav = 4.16m/s</span>
