OpG3

x_1=-60cm

E1

E2

E12

Capacitatile intre bornele 1-2 din ambele montaje trebuie sa fie egale. La fel si intre bornele 2-3 respectiv 1-3.

Rezulta relatiile:

E4-3

E4-4

E4-5

E16. Circuitele 7a si 7b se pot desena altfel:

E19

E40-1

E50-1

E50-2

E50-3

E50-4

E50-5

E50-6

E50-23

E5

E5-1

E5-2

E4.8

E4-9

E4-10

E4-11

E4-12

E4-13

E17

E18 Folosim lg. Gauss

Ca si cum toata sarcina s-ar afla in centrul sferei.

E20

E21

E=0

E40-2

E40-3

E40-4

E22.

Ref: www.oradefizica.ro

E23

E30

E30-1

E30-2

E30-3 

Problema se reduce la problema E30-2

E30-4

E30-5

E30-6

E30-7

E25-1.

F=Bil=o,6N

E26.

F=o.1N

E27. 

B=0.08T

E28.

   B=μI/2πr      B=0T; B=2.4x10^-4T

E29.

dist=1,5 cm fata de conductorul 1

F=1,33x10^-5N/cm

E.x3

N1.1

N1.3

N1.2

N1.4

N1.5

N1.6

N1.7

N1.x1

Felect = k • Q1 • Q2 / d2

Felect = (9.0 x 109 N•m2/C2) • (1.00 C) • (1.00 C) / (1.00 m)2

Felect = 9.0 x 109 N

The force of repulsion of two +1.00 Coulomb charges held 1.00 meter apart is 9 billion Newton. This is an incredibly large force that compares in magnitude to the weight of more than 2000 jetliners.

N1.x2

Felect = k • Q1 • Q2 / d2

Felect = (9.0 x 109 N•m2/C2) • (6.25 x 10-9 C) • (6.25 x 10-9 C) / (0.617 m)2

Felect = 9.23 x 10-7 N

N1.x7

Answer: 0.030 m or 3.0 cm

Step 1: Identify known values in variable form.

Q1 = 1.0 x 10-6 C and Q2 = 1.0 x 10-6 C

Felect = Fgrav = mg = 1.0 • 9.8 m/s/s = 9.8 N

N1.x3

Felect = k • Q1 • Q2 / d2

d2 • Felect = k • Q1 • Q2

d2 = k • Q1 • Q2 / Felect

d = SQRT(k • Q1 • Q2) / Felect

d = SQRT [(9.0 x 109 N•m2/C2) • (-8.21 x 10-6 C) • (+3.37 x 10-6 C) / (-0.0626 N)]

d = Sqrt [ +3.98 m2 ]

d = +1.99 m

N1.x4

Answer: E

In the equation Felect = k • Q1 • Q2 / d2 , the symbol Felect represents the electrostatic force of attraction or repulsion between objects 1 and 2. The symbol k is Coulomb's law constant (9 x 109 N • m2 / C2), Q1 and Q2 represent the quantity of charge on object 1 and object 2, and d represents the separation distance between the objects' centers.

N1.x5

Answer: 2.16 x 10-5 N, attractive

Step 1: Identify known values in variable form.

Q1 = +3.5 x 10-8 C and Q2 = -2.9 x 10-8 C

d = 0.65 m

Step 2: Identify the requested value

F = ???

N1.x7

N1.x9

Answer: 0.320 N

Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of both objects is doubled, then the force will become four times greater. Four times 0.080 N is 0.320 N.

N1.x10

Answer: 0.020 N

Explanation: The electrostatic force is inversely related to the square of the separation distance. So if d is two times larger, then F is four times smaller - that is, one-fourth the original value. One-fourth of 0.080 N is 0.020 N.

N1.x12

Answer: 0.0050 N

Explanation: The electrostatic force is inversely related to the square of the separation distance. So if d is four times larger (quadrupled), then F is 16 times smaller - that is, 1/16-th the original value. One-sixteenth of 0.080 N is 0.0050 N.

N1.x6

Answer: 1.3 x 10-2 N, attractive  (rounded from 1.296  x 10-2 N)

Step 1: Identify known values in variable form.

Q1 = +6.0 x 10-7 C and Q2 = -6.0 x 10-7 C

d = 0.50 m

Step 2: Identify the requested value

F = ???

N1.x8

Answer: 0.160 N

Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater. Two times 0.080 N is 0.160 N.

N1.x11

Answer: 0.00889 N

Explanation: The electrostatic force is inversely related to the square of the separation distance. So if d is three times larger, then F is nine times smaller - that is, one-ninth the original value. One-ninth of 0.080 N is 0.00889 N.

N1.x13

Answer: FALSE

Contrary to a commonly held belief, a lightning rod does not serve to prevent a lightning bolt. The presence of the rod on the building can only serve to divert the charge in the bolt to the ground through a low resistance pathway and thus protect the building from the damage which would otherwise result.

N1.x14

Answer: False

The presence of an elevated lightning rod could serve to draw charge from the cloud to the ground. In the event of a lightning strike, a bolt would likely select a path from the cloud that ultimately connects to the rod. If the rod is not grounded, then the charge would likely pass through the home during its journey to the ground. The intense heat associated with the lightning bolt would cause severe damage and possibly cause an explosion or a fire. In the end, it would have been better to not even have installed a lightning rod than to have installed one that is not grounded.

N1.x16

Answer: 3.83 meters

Begin by determining the Q of the balloons.

Q1 = Q2 = # of excess electrons • Qelectron = 4.0 x 10-6 C.

The force of gravity of the balloons is m • g or 0.0098 N.

To levitate the top balloon over the bottom balloon, the electrical force of repulsion must equal the force of gravity on the top balloon. Thus Felect = 0.0098 N.

Now that Q1, Q2, and F are known, Coulomb's law can be used to determine the distance d. Algebraic rearrangement leads to d = Sqrt [ (k• Q1 • Q2 ) / F ]. Substitution leads to the answer.

N1.x18

N1.x19

N1.x15

Answer: 0.052 meters

Given:

Q1 = +8.5 x 10-8 C

Q2 = -6.3 x 10-8 C

m = 0.0018 kg

Use the mass to determine the force of gravity (m • g). The force of gravity on the balloon is 0.0176 N. Thus, the upward electrical force is 0.0176 N. Now that F, Q1, and Q2 are known, Coulomb's law can be used to determine the distance d in the equation. Algebraic rearrangement leads to d = Sqrt [ (k• Q1• Q2) / F ]. Substitution leads to the answer.

N1.x17

Answer: x = -267 cm

Some reasoning would lead to the conclusion that C must be located to the left of A so that the repulsion interaction with B is balanced by the attractive interaction with A. Thus, the distance from A to C can be called x and the distance from B to C can be called 0.6 + x (where x is the absolute value of the coordinate position (in meters).

Expressions for FAC and FBC can be written

FAC = k • QA • QC / (dAC)2FBC = k • QB • QC / (dBC)2

and set equal to each other since objects at equilibrium have balanced forces. Thus

k • QA • QC / (dAC)2 = k • QB • QC / (dBC)2

The equation can be simplified by canceling k and QC. Thus,

QA / (dAC)2 = QB / (dBC)2

Substitute x and 0.6 + x into this equation:

Then solve for x by taking the square root of each side and substituting the Q values into the equatio

N1.x21

N1.x28

d = +1.99 m

N1.x30

Q1 = 1.0 x 10-6 C and Q2 = 1.0 x 10-6 C Felect = Fgrav = mg = 1.0 • 9.8 m/s/s = 9.8 N

N1.x38

N1.x39.

N1.x40

F=40N

N1.x41

F= 311.5 N

N1,x42

Forta se reduce de 4 ori

N1.x44

q3/q2=9/4

Ex10

θ=39 grade

M1-1

M1-4

M1-5

M1-6

M2

Punctul de intilnire=la 104 km fata de B

M4-2

M4-3

t=2l/(v1+v2)

M4-4

t=2l/(v2-v1)=0,0066h

M5-2

M7

M7-1

M7-2

M7-3

M7-4

M7-5

M7-6

M7.7

M7-8

M7-9

M11

M12

M13

M19

M20

M20-1

M21

Nu este corect. In sistemele de referinta neinertiale legile lui Newton nu se aplica.

M23

M23-1

M24-1

F=mg/2

M24-2

F=mg/3

M24-3

F=mg/4

M24-5

M24-6

M26

L=0.248m

M26-1bis

Pendulul A

M26-1

Osc5

Osc8

Osc9

Osc6

Osc10

Osc11

M26-5

M8.

M9

F=1718,4N

M10

M10-1

M10-2

M10-2rez.  Sfera  goală se înlocuiește un punct  de masă M șituat în centrul ei(vezi fig M10-2).

M10-3

M10-3.  Momentul forței care acționează asupra Pământului este 0. Aceasta înseamnă că dL/dt=0.

M10-4

d=Rx,

R=d/x=6800km, aproape bine

M10-6

M10-7

M14

M14-4

M15

M16

M17-1

M22

Calatorul vede mingea ca se misca pe verticala. Din sistemul legat de pamint mingea se misca dupa o parabola-din cauza vitezei vagonului.

M24

M25

M26-3

M.x5

a=0.9m/s^2

N1.8

Un1

Und2

T=7secunde

Und3

ν=0.25 Hz
T= 4.0 s

Und4

v=410 mi/hr=180 m/s

Und5

t=157 secunde

Und6

t=0,0725 s

M.x4

d=79.7m

h1

h2

h3

h4

h5

h6

h7

h8

h9

h10

L1

S5

S1