Acurve of radius 78m is banked for design speed of 85km/h.if coeficient of static friction is 0.3[wet pavement]at what range of speeds can a car safelymake the curve?
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Ответ:
The speed of the car safely make the curve, V = 65.23 Km/h
Explanation:
Given,
The radius of the curve, r = 78 m
The designed speed of the curve, v = 85 km/h
= 23.61 m/s
The coefficient of static friction, μₓ = 0.3
The static friction of the curve is given by the relation,
Fₓ = μₓ η
The acceleration responsible for the static friction
aₓ = μₓ x g
Substituting the values in the equation
aₓ = 0.3 x 9.8
= 2.94 m/s²
The designed acceleration of the curve
a₀ = v²/r
= 23.61²/78
= 7.15 m/s²
The acceleration supported by the static friction, aₓ can be subtracted from the designed acceleration of the curve.
Therefore the net acceleration,
a = a₀ - aₓ
= 7.15 - 2.94
= 4.21 m/s²
The centripetal velocity associated with this acceleration is
a = V²/r
∴ V² = a x r
V = √(a x r)
=√ (4.21 x 78)
= 18.12 m/s
= 65.23 Km/h
Hence, the speed required by the car to safely make the curve is, V = 65.23 Km/h
Ответ:
Responder: 480 vatios
Explicación:
La potencia en un circuito eléctrico am puede expresarse como el producto si la corriente y el voltaje en el circuito.
Potencia = corriente × voltaje
Corriente dada = 4 amperios
Voltaje = 120 voltios
Potencia gastada en el circuito = 4 × 120
Potencia gastada en el circuito =
480 vatios