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Alenkinab [10]
3 years ago
5

A car and a large truck traveling at the same speed collide head-on and stick together. Which vehicle experiences the larger cha

nge in the magnitude of its momentum
Physics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

Both vehicles experience the same change in momentum

Explanation:

Let m represent the mass of the vehicle, and 2m represent the mass of the large truck, and let v represent their initial speed, we have;

The total initial momentum, p_i given as follows;

p_i = 2·m·v - m·v = m·v

The total final momentum, p_f = (2·m + m) × v_f

By the principle of conservation of linear momentum, the total initial momentum = The total final momentum

m·v = (2·m + m) × v_f

m·v = 3·m·v_f

∴ v = 3 × v_f

v_f = v/3

The change in the momentum for the large truck = 2·m·v - 2·m·v_f

Therefore;

The change in the momentum for the large truck = 2·m·v - 2·m·v/3 = 2·m·(v - v/3)

The change in the momentum for the large truck = 2·m·(v - v/3) = 2·m·2·v/3 =  4·m·v/3

The change in the momentum for the car = m·v - m·(-v_f) = m·v - m·(-v)/3 = m·v + m·(v)/3 = 4·m·v/3

Therefore, the change in the momentum for the large truck = The change in the momentum for the car and both vehicles experience the same change in momentum.

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A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have
DiKsa [7]

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

\frac{Q}{V}=\frac{A\epsilon_0}{d}\\\Rightarrow \Q=V\frac{A\epsilon_0}{d}\\\Rightarrow Q=5.8\times 10^3\frac{0.024\times 8.85\times 10^{-12}}{0.0028}\\\Rightarrow Q=4.4\times 10^{-7}\ C

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

6 0
3 years ago
un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración
sergejj [24]

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en  M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

  • Vf: Velocidad final
  • Vo: Velocidad inicial
  • a: Aceleración
  • d: Distancia recorrida

En este  caso:

  • Vf: 0 m/s, porque el avión se detiene
  • Vo: 50 m/s
  • a: ?
  • d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}

a= - 10.42 m/s²

<u><em>La aceleración necesaria para detener el avión es - 10.42 m/s².</em></u>

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3 years ago
True or False:
bekas [8.4K]
I think it false. Sorry if i'm wrong.

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