Two 8V batteries are set up with four light bulbs in a complex circuit. Light bulb A has a resistance of 4.00 Ω, light bulb B ha
s a resistance of 12.00 Ω, light bulb C has a resistance of 3.00 Ω, and light bulb D has a resistance of 6.00 Ω.
***THREE SIGNIFICANT FIGURES, DO NOT TYPE THE UNIT***
Reff: Ω (Blank #1)
Veff: V (Blank #2)
Ieff: A (Blank #3)
RA: Ω (Blank #4)
VA: V (Blank #5)
IA: A (Blank #6)
RB: Ω (Blank #7)
VB: V (Blank #8)
IB: A (Blank #9)
RC: Ω (Blank #10)
VC: V (Blank #11)
IC: A (Blank #12)
RD: Ω (Blank #13)
VD: V (Blank #14)
ID: A (Blank #15)
***THREE SIGNIFICANT FIGURES, DO NOT TYPE THE UNIT***
Reff: Ω (Blank #1)
Veff: V (Blank #2)
Ieff: A (Blank #3)
RA: Ω (Blank #4)
VA: V (Blank #5)
IA: A (Blank #6)
RB: Ω (Blank #7)
VB: V (Blank #8)
IB: A (Blank #9)
RC: Ω (Blank #10)
VC: V (Blank #11)
IC: A (Blank #12)
RD: Ω (Blank #13)
VD: V (Blank #14)
ID: A (Blank #15)
1 answer:
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Answer:
Vf = 210 [m/s]
Av = 105 [m/s]
y = 2205 [m]
Explanation:
To solve this problem we must use the following formula of kinematics.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0 (released from the rest)
g = gravity acceleration = 10 [m/s²]
t = time = 21 [s]
Vf = 0 + (10*21)
Vf = 210 [m/s]
Note: The positive sign for the gravity acceleration means that the object is falling in the same direction of the gravity acceleration (downwards)
The average speed is defined as the sum of the final speed plus the initial speed divided by two. (the initial velocity is zero)
Av = (210 + 0)/2
Av = 105 [m/s]
To calculate the distance we must use the following equation of kinematics
44100 = 20*y
y = 2205 [m]
