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Pi Workgroup

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Submitted By jonber2834
Words 378
Pages 2
q = 2 C q = 2 C
∆V=100 million Volts
∆V=100 million Volts
Page 435: problem #7 V= EPEq Need to find the voltage but we do not know what EPE is. Arrange problem to solve for EPE: EPE= qv = 2C x 100,000,000 =
200,000,000 VC or 2E9J

Page 454: one-step calculations #6
P = IV
I = 10 A
V = 120 V
10 A * 120 V = 1200W
P = IV
I = 10 A
V = 120 V
10 A * 120 V = 1200W
V = 120
V = 120

exercise #9

1 2 3 4 5 1 2 3 4 5

Only #5 will work.

Page 457: problems #3
C (current) = I
P = IV to get current PI = 1200 W120 V 10 A
I = 10 A
Ohms Law
V = IR
VI = IRI = VI = R = 120 V10 A = 12 Ω
C (current) = I
P = IV to get current PI = 1200 W120 V 10 A
I = 10 A
Ohms Law
V = IR
VI = IRI = VI = R = 120 V10 A = 12 Ω
P = 1200 W
P = 1200 W

V = 120
V = 120

HW. 3 3.1
Pg. 435
6. 2x106 1.6/3.2 10-10x10-4
Pg454
4. Power= 2w
5. 0.5x120= 60v
Pg. 457
2. Resistance= 6 ohms
Pg. 435 #4 me= 9.109x10-31 kg mE= 6x1024 kg q= 1.602x10-19 C
E= 10000 vm
F= qE
F= 1.602x10-19 C x 10000 vm
F= 1.602x10-15 N
Fg= Gme mE r2 re= 6.4x106 m
G= 6.67x10-11 Nm2kg2
Fg = 6.67x10-11 Nm2kg2 x 9.109x10^-31 kg x 6x10^24 kg (6.4x106 m)2= 8.89x10-30 N

Pg. 454 #1
R= 15 Ω
V= 120 V
I=V/R
I= 120 V15 Ω= 8 A
Pg. 454 #2
R=90 Ω
V=9 V
I= V/R
I= 9 V90 Ω= .1 A
Pg. 454 #3
R=1000 Ω
V=6 V
I= V/R
I= 6 V1000 Ω= .006 A
Pg. 456 #44
The brightness of Bulb A will remain the same because the voltage of the battery will be equal to voltage of all bulbs connected.
Pg.457 #1
P= 60 W
V= 120 V
P=IV
I=P/V
I= 60 W 120 V = .5 A

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