<span>Answer:The weight of the door creates a CCW torque given by
Tccw = 145 N*3.13 m / 2
You need a CW torque that's equal to that
Tcw = F*2.5 m*sin20</span>
Answer:
El peso del cartel es 397,97 N
Explanation:
La tensión dada en cada segmento del cable = 2000 N
El desplazamiento vertical del cable = 50 cm = 0,5 m
La distancia entre los polos = 10 m
La posición del letrero en el cable = En el medio = 5
El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °
El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero
El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N
El peso del signo = 397,97 N.
Make an equation for both.
Ford= 15t+200
Chevy=20t
Now you must set them equal and solve for time t. Giving you 40 seconds. Now plug that in to the chevy equation to get 800 meters.
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).
<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>
<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
