• Vietato rispondere a domande sulla teoria della relatività?

    Best answer: ...per carnevale ogni scherzo vale ^_-
    Best answer: ...per carnevale ogni scherzo vale ^_-
    2 answers · Fisica · 10 months ago
  • Come calcolare la resistività di un materiale come il rame.?

    Best answer: R = ρ*ℓ/s ρ = R*s/ℓ occorre conoscere resistenza R, sezione s e lunghezza ℓ...per tua informazione la resistività del rame ρcu vale circa 17*10^-3 ohm*mm^2/m ed è variabile con la temperatura in accordo alla formula : R2 = R1*(234,5+T2)/(234,5+T1)
    Best answer: R = ρ*ℓ/s ρ = R*s/ℓ occorre conoscere resistenza R, sezione s e lunghezza ℓ...per tua informazione la resistività del rame ρcu vale circa 17*10^-3 ohm*mm^2/m ed è variabile con la temperatura in accordo alla formula : R2 = R1*(234,5+T2)/(234,5+T1)
    4 answers · Fisica · 4 days ago
  • Can anyone help me with my physics problem?

    Best answer: m*V^2 = k*x^2 V = √k*x^2/m = √100*0.2^2/2 = √100*(4/100)/2 = √2 m/sec
    Best answer: m*V^2 = k*x^2 V = √k*x^2/m = √100*0.2^2/2 = √100*(4/100)/2 = √2 m/sec
    1 answer · Physics · 3 weeks ago
  • Can anyone help me with my scalar product problem?

    Best answer: V1 = 10 V2 = 11*0,909 arccos 0,909 = 24.62° sin 24.62° = 0.4166 A+B = 11*sin 24.62 = 4.58
    Best answer: V1 = 10 V2 = 11*0,909 arccos 0,909 = 24.62° sin 24.62° = 0.4166 A+B = 11*sin 24.62 = 4.58
    2 answers · Physics · 3 weeks ago
  • A) find the work done by the 94N force. B) Find the magnitude of the work done by the coefficient or frciction answer is in Joules?

    Best answer: A) work = F*cos 25.4*d = 94*51.5*0.9033 = 4.373,0 joule B) work = (m*g-94*sin 25.4)*0.19*51.5 = (18.2*9.806-94*0.429)*0.19*51.5 = 1352 joule
    Best answer: A) work = F*cos 25.4*d = 94*51.5*0.9033 = 4.373,0 joule B) work = (m*g-94*sin 25.4)*0.19*51.5 = (18.2*9.806-94*0.429)*0.19*51.5 = 1352 joule
    1 answer · Physics · 2 days ago
  • Capacitor charge and discharge question. I just don't get it... Can anyone please help? Thanks in advance?

    Best answer: A 68 uf capacitor is charged to a p.d of 9.0 V then discharged through a 20 kilo ohms resistor. Calculate: i) The charge stored by the capacitor at a p.d. of 9.0 V Q = C*V = 68*9*10^-6 = 612 μCoulomb ii) The initial discharge current Io = V/R = 9*10^6/(20*10^3) = 9000/20 = 450 μA b) Calculate the p.d. and the discharge current... show more
    Best answer: A 68 uf capacitor is charged to a p.d of 9.0 V then discharged through a 20 kilo ohms resistor. Calculate: i) The charge stored by the capacitor at a p.d. of 9.0 V Q = C*V = 68*9*10^-6 = 612 μCoulomb ii) The initial discharge current Io = V/R = 9*10^6/(20*10^3) = 9000/20 = 450 μA b) Calculate the p.d. and the discharge current 5.0 s after the discharge started Ƭ = R*C = 68*20*10^3*10^-6 = 1.36 sec V = Vo*e^-t/Ƭ = 9*2.7182818^(-5/1.36) = 0.2278 V I = V/R = 227.8*10^3/(20*10^3) = 11.39 μA
    2 answers · Physics · 3 days ago
  • A sled starts from rest and slides 10.0 m down a 28.0 degree frictionless hill. What is its acceleration?

    Best answer: a = g*sin 28 = 9.806*0.469 = 4.604 m/sec^2
    Best answer: a = g*sin 28 = 9.806*0.469 = 4.604 m/sec^2
    3 answers · Physics · 5 days ago
  • Mi potete risolvere questo problema???????????? Plsssss?

    Best answer: pressione p (N/m^2) = F/A = 24*10^4/1.8 p = ρa*g*h (ρa essendo la densità dell'acqua pari a 1000 kg/m^3) h = p/(ρa*g) = 24*10^4/(1.8*10^3*9,806) = 24*10/(1,8*9,806) = 13,60 m
    Best answer: pressione p (N/m^2) = F/A = 24*10^4/1.8 p = ρa*g*h (ρa essendo la densità dell'acqua pari a 1000 kg/m^3) h = p/(ρa*g) = 24*10^4/(1.8*10^3*9,806) = 24*10/(1,8*9,806) = 13,60 m
    4 answers · Fisica · 2 weeks ago
  • Il lato di una pedana quadrata = 2,5 m ed errore assoluto = 0,1. calcola l'area della pedana e l'errore in percentuale. grazie mille!?

    Best answer: area base A = 2,5^2 = 6,25 m^2 area con errore A' = (2,5+0,1)^2 = 2,6^2 = 6,76 m^2 errore percentuale ε% = 100*(6,76-6,25)/6,25 = 8,16%
    Best answer: area base A = 2,5^2 = 6,25 m^2 area con errore A' = (2,5+0,1)^2 = 2,6^2 = 6,76 m^2 errore percentuale ε% = 100*(6,76-6,25)/6,25 = 8,16%
    1 answer · Fisica · 1 week ago
  • Aiuto! Non riesco a risolvere questi due problemi di fisica legati al suono :(?

    Best answer: P = 0,5 watt calcolo superfici sfere corrispondenti : A1 = PI*60^2 A2 = PI*120^2 A3 = PI*240^2 A4 = PI*300^2 riferimento Iref scala logaritmica intensità sonora Ii = 10^-12 watt/m^2 dB = 10log(Ii/Iref) I1 = 10 log ( 0,5/3,1416*60^2)/10^-12 = 10 log 5*10^11/(10^3*3,6*3,1416) = 76,46 dB I2 = 10 log ( 0,5/3,1416*120^2)/10^-12 = 10 log... show more
    Best answer: P = 0,5 watt calcolo superfici sfere corrispondenti : A1 = PI*60^2 A2 = PI*120^2 A3 = PI*240^2 A4 = PI*300^2 riferimento Iref scala logaritmica intensità sonora Ii = 10^-12 watt/m^2 dB = 10log(Ii/Iref) I1 = 10 log ( 0,5/3,1416*60^2)/10^-12 = 10 log 5*10^11/(10^3*3,6*3,1416) = 76,46 dB I2 = 10 log ( 0,5/3,1416*120^2)/10^-12 = 10 log 5*10^11/(10^3*14,4*3,1416) = 70,43 dB I3 = 10 log ( 0,5/3,1416*240^2)/10^-12 = 10 log 5*10^11/(10^3*57,6*3,1416) = 64,41 dB I4 = 10 log ( 0,5/3,1416*300^2)/10^-12 = 10 log 5*10^11/(10^3*90,0*3,1416) = 62,48 dB Quale sarà la distanza minima di osservanza per le nuove abitazioni ? poco meno di 120 m Sono barriere fonoassorbenti o schermanti atte a ridurre a limiti accettabili l'intensità sonora percepita da abitazioni prossime a sorgenti sonore quali autoveicoli e/o treni
    2 answers · Fisica · 1 week ago
  • Derive the formula for the escape veolocity of a projectile from a plant of mass M?

    Best answer: Escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero. An object which has achieved escape velocity can't be either on the surface, or in a closed orbit of any radius; escape velocity in a direction pointing away from the ground of a massive body, will... show more
    Best answer: Escape velocity is the speed at which the sum of an object's kinetic energy and its gravitational potential energy is equal to zero. An object which has achieved escape velocity can't be either on the surface, or in a closed orbit of any radius; escape velocity in a direction pointing away from the ground of a massive body, will make the object moving away from the body, slowing forever and approaching, but never reaching, zero speed. Once escape velocity is achieved, no further impulse is needed for going on in its escape : in other words, with escape velocity the object will move away from the other massive body, continually slowing, and will asymptotically approach zero speed as the mutual distance approaches infinity, and will never come back. Speeds higher than escape velocity have a positive speed at infinity ; furthermore the minimum escape velocity is evaluated on the assumption that there is no friction (e.g., atmospheric drag which would increase the required instantaneous velocity to escape the gravitational influence) and that there will be no further sources of additional energy thrusters), which would reduce the required instantaneous velocity. For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula : Ve = √2 G M/ R where : G is the universal gravitational constant (G ≈ 6.67×10−11 m^3·kg−1·s^−2) M the mass of the body to be escaped from R the distance from the center of mass of the body to the object. The equation shows that Ve is independent of the mass of the escaping object
    4 answers · Physics · 6 months ago
  • It is a consequence of Newton's law of gravitation that near the surface of any planet,?

    Best answer: b) The distance fallen in 12 seconds is 4.00 times the distance fallen in 6 seconds. (2t)^2 =4t^2
    Best answer: b) The distance fallen in 12 seconds is 4.00 times the distance fallen in 6 seconds. (2t)^2 =4t^2
    1 answer · Physics · 1 week ago
  • Problema fisica please?

    Best answer: mentre inizia a muoversi : forza di attrito Fr = m*g*μs = 3*9,806*0,5 = 14,71 N forza accelerante Fa = m*a = 3*2,2 = 6,60 N forza totale F = Fr+Fa = 14,71+6,60 = 21,31 N in movimento : accelerazione a = (21,31-3*9,806*0,46)/3 = 2,59 m/sec^2 forza di attrito Fr = m*g*μs = 3*9,806*0,5 = 14,71 N emiforza F' = F/2 = 21,31/2 = 10.65 N... show more
    Best answer: mentre inizia a muoversi : forza di attrito Fr = m*g*μs = 3*9,806*0,5 = 14,71 N forza accelerante Fa = m*a = 3*2,2 = 6,60 N forza totale F = Fr+Fa = 14,71+6,60 = 21,31 N in movimento : accelerazione a = (21,31-3*9,806*0,46)/3 = 2,59 m/sec^2 forza di attrito Fr = m*g*μs = 3*9,806*0,5 = 14,71 N emiforza F' = F/2 = 21,31/2 = 10.65 N < 14,71 N .....il corpo non si muove
    3 answers · Fisica · 1 week ago
  • An automobile is moving at speed of 2.00 m/s along a circular road of radius 20 m?

    Best answer: An automobile is moving at speed of 2.00 m/s along a circular road of radius 20 m A) Find the time it takes to make one full circle when the car is still moving at constant speed. t = C/V = 2*20*π/2 = 20π sec ( ≈ 62.83 sec) B) Find the car radial acceleration during this time. ac = V^2/r = 2^2/20 = 4/20 = 0.20 m/sec^2 C) Then its... show more
    Best answer: An automobile is moving at speed of 2.00 m/s along a circular road of radius 20 m A) Find the time it takes to make one full circle when the car is still moving at constant speed. t = C/V = 2*20*π/2 = 20π sec ( ≈ 62.83 sec) B) Find the car radial acceleration during this time. ac = V^2/r = 2^2/20 = 4/20 = 0.20 m/sec^2 C) Then its speed starts increasing at a rate of 0.600 m/s2 while it stays on the same road. When the instantaneous speed of the automobile is 4.00 m/s, how long (t) would it take for the car to make one full turn from the moment it started its tangential acceleration? What you are asking can be understood with different meanings : a) the car increases it speed fom 2 to 4 m/sec , then the speed remains constant for the rest of the turn acceleration time ta = (Vf-Vi)/a = (4-2)/0,6 = 20/6 = 10/3 sec C = 2*20*π = (Vi+Vf)/2*ta+Vf*tc time at constant speed tc = (40*3.1416-3*10/3)/4 = (125.6-10)/4 = 115.6/4 = 28.9 sec t = ta+tc = 28.9+10/3 = 32.23 sec b) the car stretches a full turn @ 4 m/sec t = 40*π/4 = 10π = 31.41sec c) ) the car stretches a full turn @ constant acceleration 40*π = Vi*t+a*t*t/2 125.6-2t-0.3*t^2 = 0 t = (2-√2^2+1.2*125.6)/-0.6 = 14.40 sec
    1 answer · Physics · 1 week ago
  • A descending elevator of mass 740 kg is uniformly decelerated to rest over a distance of 3 m by a cable in which the tension is 9288 N.?

    Best answer: (0-(Vi^2)) = 2*a*d a = -(T/m-g) = -(9288/740-9.8) = -2.75 m/sec^2 Vi = √2*3*2.75 = 4.06 m/sec
    Best answer: (0-(Vi^2)) = 2*a*d a = -(T/m-g) = -(9288/740-9.8) = -2.75 m/sec^2 Vi = √2*3*2.75 = 4.06 m/sec
    2 answers · Physics · 3 weeks ago
  • Help please with Physics problem dealing with rotation?

    Best answer: Suppose a piece of dust rests on a CD. n = 460 rpm r = 4.8 cm t = 2.5 min a) If the spin rate of the CD is n = 460 rpm what distance N (in radians) is traveled by the piece of dust in 2.5 mins? N = 2PI*n*t = 460*2.5*2*PI = 4600PI/2 = 2300PI radians b) If the dust is 4.8 cm away from the center, what is the total distance d... show more
    Best answer: Suppose a piece of dust rests on a CD. n = 460 rpm r = 4.8 cm t = 2.5 min a) If the spin rate of the CD is n = 460 rpm what distance N (in radians) is traveled by the piece of dust in 2.5 mins? N = 2PI*n*t = 460*2.5*2*PI = 4600PI/2 = 2300PI radians b) If the dust is 4.8 cm away from the center, what is the total distance d traveled by it during the same time interval in meters? d = V*t = ω*r*t = 2PI*n/60*0.048*(2.5*60) = 0.10472*460*0.048*150 = 346.833 m
    1 answer · Physics · 1 week ago
  • A 27-g rifle bullet traveling 180 m/s embeds itself in a 3.4-kg pendulum hanging on a 2.6-m-long string,?

    Best answer: A 27-g rifle bullet traveling 180 m/s embeds itself in a 3.4-kg pendulum hanging on a 2.6-m-long string which makes the pendulum swing upward in an arc. a) Determine the vertical component Δy of the pendulum's maximum displacement. b) Determine the horizontal component Δx of the pendulum's maximum displacement. Momentum I =... show more
    Best answer: A 27-g rifle bullet traveling 180 m/s embeds itself in a 3.4-kg pendulum hanging on a 2.6-m-long string which makes the pendulum swing upward in an arc. a) Determine the vertical component Δy of the pendulum's maximum displacement. b) Determine the horizontal component Δx of the pendulum's maximum displacement. Momentum I = mb*Vb = 27*10^-3*180 = 4.860 kg*m/sec Vp = I/(mb+mp) = 4.860/(0.027+3.4) = 1.4181 m/sec Δy = Vp^2/2g = 1.4181^2/19.612 = 0.1025 m = 2.6*(1-cosΘ) (1-cos Θ) = 0.1025/2.6 = 0.0394 cos Θ = 1-0.0394 = 0.9606 sin Θ = √1-0.9606^2) = 0.2781 Δx = 2.6*sin Θ = 0.2781*2.6 = 0.723 m
    2 answers · Physics · 1 week ago
  • FISICA!!!!?

    Best answer: Ux = 8,1*sen 10° = 1,407 Uy = -8,1*cos 10° = -7,977 Wx = -1,5*sen 15° = -0,388 Wy = 1,5*cos 15° = 1,449 Rx = Ux+Wx = 1,407-0,388 = 1,018 Ry = Uy+Wy = -7,977+1,449 = -6,528 R = √1,018^2+6,528^2 = 6,607
    Best answer: Ux = 8,1*sen 10° = 1,407 Uy = -8,1*cos 10° = -7,977 Wx = -1,5*sen 15° = -0,388 Wy = 1,5*cos 15° = 1,449 Rx = Ux+Wx = 1,407-0,388 = 1,018 Ry = Uy+Wy = -7,977+1,449 = -6,528 R = √1,018^2+6,528^2 = 6,607
    2 answers · Fisica · 2 weeks ago
  • Se in una pentola in acciaio inox senza coperchio?

    Best answer: Dell'acqua si, della pentola no !!!
    Best answer: Dell'acqua si, della pentola no !!!
    1 answer · Fisica · 2 weeks ago