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Pentru enunturi clic AICI
OPTICA
Intrebari:
I1=4, I2=2, I3=2, I4=3, I5=4, I6=2, I7=3,I8=2, I9=1, I10=3, I11=1,I12=2, I13=3, I14=2
ELECTRICITATE
E1 
E2 
E12 Capacitatile intre bornele 12 din ambele montaje trebuie sa fie egale. La fel si intre bornele 23 respectiv 13. Rezulta relatiile: 
E43 E44 E45 
E16. Circuitele 7a si 7b se pot desena altfel: E19 E401 E501 E502 E503 E504 E505 E506 E5023 
E5 E51 E52 E4.8 E49 E410 E411 E412 E413 E17 E18 Folosim lg. Gauss Ca si cum toata sarcina sar afla in centrul sferei. E20 E21 E=0 E402 E403 E404 
E22. Ref: www.oradefizica.ro 
E23 E30 E301 E302 E303 Problema se reduce la problema E302 E304 E305 E306 E307 E251. 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 109 C) • (6.25 x 109 C) / (0.617 m)2 Felect = 9.23 x 107 N N1.x7 Answer: 0.030 m or 3.0 cm Step 1: Identify known values in variable form. Q1 = 1.0 x 106 C and Q2 = 1.0 x 106 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 106 C) • (+3.37 x 106 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 105 N, attractive Step 1: Identify known values in variable form. Q1 = +3.5 x 108 C and Q2 = 2.9 x 108 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, onefourth the original value. Onefourth 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, 1/16th the original value. Onesixteenth of 0.080 N is 0.0050 N. 
N1.x6 Answer: 1.3 x 102 N, attractive (rounded from 1.296 x 102 N) Step 1: Identify known values in variable form. Q1 = +6.0 x 107 C and Q2 = 6.0 x 107 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, oneninth the original value. Oneninth 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 106 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 108 C Q2 = 6.3 x 108 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: QA / x2 = QB / ( 0.6 + x )2Then 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 106 C and Q2 = 1.0 x 106 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 
MECANICA
M11 M14 M15 M16 
M2 Punctul de intilnire=la 104 km fata de B M42 M43 t=2l/(v1+v2) 
M44  
t=2l/(v2v1)=0,0066h  
M52  
M7 M71 M72 M73 M74 M75 M76 M7.7 M78 M79 M11 M12 M13 M19 M20 M201 M21 Nu este corect. In sistemele de referinta neinertiale legile lui Newton nu se aplica. M23 M231 M241 F=mg/2 M242 F=mg/3 M243 F=mg/4 M245 M246 M26 L=0.248m M261bis Pendulul A M261 Osc5 Osc8 Osc9 Osc6 Osc10 Osc11 M265 
M8. M9 F=1718,4N M10 M101 M102 M102rez. Sfera goală se înlocuiește un punct de masă M șituat în centrul ei(vezi fig M102). M103 M103. Momentul forței care acționează asupra Pământului este 0. Aceasta înseamnă că dL/dt=0. M104 d=Rx, R=d/x=6800km, aproape bine M106 M107 M14 M144 M15 M16 M171 M22 Calatorul vede mingea ca se misca pe verticala. Din sistemul legat de pamint mingea se misca dupa o paraboladin cauza vitezei vagonului. M24 M25 M263 M.x5 a=0.9m/s^2 N1.8 Un1 Und2 T=7secunde Und3 ν=0.25 Hz 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 
SATELITI
S5 S1 