Here you go. I hope this will help
9514 1404 393
Answer:
11.3 units
Step-by-step explanation:
The distance can be found using the distance formula:
d = √((x2 -x1)² +(y2 -y1)²)
d = √((6 -(-2))² +(-4-4)²) = √(8² +(-8)²) = √(64·2)
d = 8√2 ≈ 11.3
The distance is approximately 11.3 units.
Answer:
B) Falce
Step-by-step explanation:
Cylinder base is circle
a.
×
A
b.
×
A
c.
A
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
<h3>How to determine the current</h3>
The formula for finding current
I = V/R
Where v = voltage
R = resistance
A. V = 12V
R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series
![I = \frac{12}{19700}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B12%7D%7B19700%7D)
×
A
B. V = 9V
R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series
![I = \frac{12}{4700}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B12%7D%7B4700%7D)
×
A
C. V= 12V
1/R =
=
× ![10^-3](https://tex.z-dn.net/?f=10%5E-3)
R =
= 310. 56 Ω
![I = \frac{12}{310. 56}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B12%7D%7B310.%2056%7D)
A
It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
Learn more about Ohms law here:
brainly.com/question/14296509
#SPJ1
Answer:
Ace is the next level after expert.