• Braking force, physcis?

    Best answer: If We call : F1 the gravitational force pulling her down Ff the friction force slowing her down and with versus opposite to F1 m the mass of the skier a the acceleration F1-Ff the accelerating force = m*a ..equation to use il quite simple : m*a = F1-Ff 75*4 = 550-Ff Ff = 550-300 = 250 N
    Best answer: If We call : F1 the gravitational force pulling her down Ff the friction force slowing her down and with versus opposite to F1 m the mass of the skier a the acceleration F1-Ff the accelerating force = m*a ..equation to use il quite simple : m*a = F1-Ff 75*4 = 550-Ff Ff = 550-300 = 250 N
    2 answers · Physics · 1 day ago
  • Find velocity and acceleration at point B in respect to Point A?

    Best answer: point A tangential speed Va = 35 m/sec tangential acceleration ata = 1.5 m/sec^2 centripetal acceleration aca = Va^2/ra = 35^2/200 = 6.125 m/sec^2 total acceleration aa = √6.125^2+1.5^2 = 6.306 m/sec*2 acceleration heading = arctan 1.5/-6.125 = 13.76° NoW point B tangential speed Vb = -25 m/sec tangential acceleration atb = 0... show more
    Best answer: point A tangential speed Va = 35 m/sec tangential acceleration ata = 1.5 m/sec^2 centripetal acceleration aca = Va^2/ra = 35^2/200 = 6.125 m/sec^2 total acceleration aa = √6.125^2+1.5^2 = 6.306 m/sec*2 acceleration heading = arctan 1.5/-6.125 = 13.76° NoW point B tangential speed Vb = -25 m/sec tangential acceleration atb = 0 m/sec^2 centripetal acceleration acb = Vb^2/rb = 25^2/100 = 6.25 m/sec^2 total acceleration ab = √6.25^2+0 = 6.25 m/sec*2 due East Va-Vb = 35-(-25) = 60 m/sec aa-ab = √(aca+acb)^2+ata^2 = √(6.125+6.25)^2+1.5^2 = 12.47 m/sec^2
    1 answer · Physics · 2 days ago
  • PHYSICS HELP!!!!! A cheetah can run at a maximum speed of 103 km/ h and a ....?

    Best answer: 75.9/3.6*t +77.4 = 103/3.6*t (103-75.9)*t = 77.4*3.6 t = 77.4*3.6/27.10 = 10.282 sec or 77.4 = (103-75.9)/3.6*t = 7.528t t = 77.4/7.528 = 10.282 sec
    Best answer: 75.9/3.6*t +77.4 = 103/3.6*t (103-75.9)*t = 77.4*3.6 t = 77.4*3.6/27.10 = 10.282 sec or 77.4 = (103-75.9)/3.6*t = 7.528t t = 77.4/7.528 = 10.282 sec
    3 answers · Physics · 4 days ago
  • Need help with this physics problem?

    Best answer: horizontal distance Lx = L*cos Θ = 19.7*cos 16.5 = 18.89 m vertical distance Ly = L*sin Θ = 19.7*sin 16.5 = 5.60 m
    Best answer: horizontal distance Lx = L*cos Θ = 19.7*cos 16.5 = 18.89 m vertical distance Ly = L*sin Θ = 19.7*sin 16.5 = 5.60 m
    1 answer · Physics · 5 days ago
  • Is there a better place than a government school to indoctrinate children ???

    Best answer: Yessss....certain "families" ....
    Best answer: Yessss....certain "families" ....
    6 answers · Physics · 5 days ago
  • Physics Question?

    Best answer: total energy E = m*g*(h+d) = 0.14*9.806*(2.8+0.062) stopping work W = E = F*d F = E/d = 0.14*9.806*(2.8+0.062)*1000/62 = 63.37 N
    Best answer: total energy E = m*g*(h+d) = 0.14*9.806*(2.8+0.062) stopping work W = E = F*d F = E/d = 0.14*9.806*(2.8+0.062)*1000/62 = 63.37 N
    1 answer · Physics · 5 days ago
  • Free fall/air resistance?

    Best answer: Air resistance implies that a limit speed will take place, after that falling speed won't increase any longer As a consequence , speed difference ΔV will increase at the beginning, then will remain constant , as well as the distance
    Best answer: Air resistance implies that a limit speed will take place, after that falling speed won't increase any longer As a consequence , speed difference ΔV will increase at the beginning, then will remain constant , as well as the distance
    1 answer · Physics · 5 days ago
  • Are dark matter and dark energy still one of the biggest Unknowns in physics, or else it won’t be called “dark” anymore?

    Best answer: They ought to be called darker and darker...
    Best answer: They ought to be called darker and darker...
    2 answers · Physics · 1 week ago
  • Please help on this one....?

    Best answer: work W = F*cos 30°*d = 85*0.866*12 = 883.3 joule
    Best answer: work W = F*cos 30°*d = 85*0.866*12 = 883.3 joule
    2 answers · Physics · 6 days ago
  • If the jet comes to rest in 13.5 s, what is the magnitude of it’s average acceleration? What is the direction of its average acceleration?

    Best answer: a aver = ΔV/Δt = (0-120)/13.5 = -8.(8) m/sec^2
    Best answer: a aver = ΔV/Δt = (0-120)/13.5 = -8.(8) m/sec^2
    2 answers · Physics · 6 days ago
  • Physics Help Please?

    Best answer: a) Vector A has length of 7 meters and is directed at 78 degrees above the -x axis. What is its y component? Ay = 7*sin 78° = 7*0.978 = 6.847 m Ax = -7*cos 78° = -7*0.208 = -1.455 b) Vector A has length of 7 meters and is directed at 21 degrees above the -x axis. What is its x component? Ax = -7*cos 21° = -7*0.934 = -6.535 Ay = 7*sin... show more
    Best answer: a) Vector A has length of 7 meters and is directed at 78 degrees above the -x axis. What is its y component? Ay = 7*sin 78° = 7*0.978 = 6.847 m Ax = -7*cos 78° = -7*0.208 = -1.455 b) Vector A has length of 7 meters and is directed at 21 degrees above the -x axis. What is its x component? Ax = -7*cos 21° = -7*0.934 = -6.535 Ay = 7*sin 21° = 7*0.358 = 2.509 c) Vector A has length of 8 meters and is directed at 56 degrees below the +x axis. What is its y component? Ay = -8*sin 56 = -8*0.829 = -6.632 m Ax = 8*cos 56 = 8*0.559 = 4.474 m
    2 answers · Physics · 1 week ago
  • A ball is thrown vertically upward with a speed of 26.0 m/s.?

    Best answer: A ball is thrown vertically upward with a speed of 26.0 m/s.? (a) How high h in meters does it rise? h = Vo^2/2g = 26^2/19.612 = 34.47 m (b) How long t in sec does it take to reach its highest point? t = Vo/g = 26/9.806 = 2.651 sec (c) How long t1 in sec does the ball take to hit the ground after it reaches its highest point? t1 = t... show more
    Best answer: A ball is thrown vertically upward with a speed of 26.0 m/s.? (a) How high h in meters does it rise? h = Vo^2/2g = 26^2/19.612 = 34.47 m (b) How long t in sec does it take to reach its highest point? t = Vo/g = 26/9.806 = 2.651 sec (c) How long t1 in sec does the ball take to hit the ground after it reaches its highest point? t1 = t = 2.651 sec (d) What is its velocity when it returns to the level from which it started? Vf = -Vo = -26.0 m/sec
    4 answers · Physics · 2 weeks ago
  • Problema di meccanica?

    Best answer: 1,5*m*r^2 = 0.5*m*r^2+m/3*r^2+m2*r^2 (9/6-5/6)*m*r^2 = m2*r^2 m2 = 2/3*m ωo = L/J = 2L/(3*m*r^2) t = 4PI/ω t = J*ω/C 4*PI*C = J*ω^2 C = J*ω^2/(4*PI)
    Best answer: 1,5*m*r^2 = 0.5*m*r^2+m/3*r^2+m2*r^2 (9/6-5/6)*m*r^2 = m2*r^2 m2 = 2/3*m ωo = L/J = 2L/(3*m*r^2) t = 4PI/ω t = J*ω/C 4*PI*C = J*ω^2 C = J*ω^2/(4*PI)
    2 answers · Fisica · 2 weeks ago
  • Physics question WITH explanation? EAST 10 PTS!!?

    Best answer: h = g/2*t^2 = -5*1^2 = -5 m ( 5 meters down)
    Best answer: h = g/2*t^2 = -5*1^2 = -5 m ( 5 meters down)
    2 answers · Physics · 1 week ago
  • Please help?

    Best answer: 37 = 250*0.145+a/2*0.145^2 74-72.5 = a*0.145^2 a = 1.5/0.145^2 = 71.34 mm/sec^2 = 7.13 cm/sec^2 = 0.0713 m/sec^2
    Best answer: 37 = 250*0.145+a/2*0.145^2 74-72.5 = a*0.145^2 a = 1.5/0.145^2 = 71.34 mm/sec^2 = 7.13 cm/sec^2 = 0.0713 m/sec^2
    1 answer · Physics · 1 week ago
  • Physics help -kinematics?

    Best answer: h = ho+Vo*t-g/2*t^2 h = 1.90+8.8*1.59-4.903*1.59^2 = 3.50 m
    Best answer: h = ho+Vo*t-g/2*t^2 h = 1.90+8.8*1.59-4.903*1.59^2 = 3.50 m
    1 answer · Physics · 1 week ago
  • Physics help!!?

    Best answer: sin 18° = 0.3090 Friction is not in the picture, therefore energy conservation shall apply !!! m*g*ℓ*sin 18° = m/2*V^2 mass m cross V = √2*g*ℓ*sin 18° = √2*9.806*259*0.3090 = 39.62 m/sec
    Best answer: sin 18° = 0.3090 Friction is not in the picture, therefore energy conservation shall apply !!! m*g*ℓ*sin 18° = m/2*V^2 mass m cross V = √2*g*ℓ*sin 18° = √2*9.806*259*0.3090 = 39.62 m/sec
    1 answer · Physics · 1 week ago
  • Aiuto in fisica?

    Best answer: Si dice aiuto come per qualsiasi altra disciplina ...^_-
    Best answer: Si dice aiuto come per qualsiasi altra disciplina ...^_-
    3 answers · Fisica · 7 months ago
  • Compare from greatest to least the accelerations of these skydivers?

    Best answer: g is approximated to 10 m/sec^2 a) 1000-N man with 800N of air drag mass m = 1000/10 = 100 kg acceleration a = (weight-drag)/mass = (1000-800)/100 = 2.0 m/sec^2 b) 800-N woman with 700N OF air drag mass m = 800/10 = 80 kg acceleration a = (weight-drag)/mass = (800-700)/80 = 1.25 m/sec^2 c)700-N woman with 600N of air drag mass m =... show more
    Best answer: g is approximated to 10 m/sec^2 a) 1000-N man with 800N of air drag mass m = 1000/10 = 100 kg acceleration a = (weight-drag)/mass = (1000-800)/100 = 2.0 m/sec^2 b) 800-N woman with 700N OF air drag mass m = 800/10 = 80 kg acceleration a = (weight-drag)/mass = (800-700)/80 = 1.25 m/sec^2 c)700-N woman with 600N of air drag mass m = 700/10 = 70 kg acceleration a = (weight-drag)/mass = (700-600)/70 = 1.43 m/sec^2 d) 100-N dog with 90 N of air drag mass m = 100/10 = 10 kg acceleration a = (weight-drag)/mass = (100-90)/10 = 1.00 m/sec^2 then a) ; c) ; b) ; d)
    4 answers · Physics · 2 weeks ago
  • Fisica Due persone trascinano una nave con delle funi che formano un angolo di 90°.F1 e F2 sono uguali e Ftot=3,8*10^4N il modulo di F1 e F2?

    Best answer: F1 = F2 = Ftot/2*√2 = (1.9*√2)*10^4 N
    Best answer: F1 = F2 = Ftot/2*√2 = (1.9*√2)*10^4 N
    1 answer · Fisica · 2 weeks ago