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Lilit [14]
2 years ago
13

Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo

si se sabe que w=2pi f
Physics
1 answer:
zubka84 [21]2 years ago
3 0

Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

ω = 10800πrad / 60 s

ω = 180πrad / seg

ω = 565.56 rad / seg

SI)

Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

C) Periodo (T)

Recordar T = 1 / f

Por lo tanto,

T = 1/90

T = 0.0111111s

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
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A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ϕ1 with
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Answer:

x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}

Explanation:

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Now, the horizontal and vertical components of these forces are:

F_{1x} = -F_1cos(\phi_1)\\F_{1y}=F_1sin(\phi_1)\\\\F_{2x}=F_2cos(\phi_2)\\F_{2y}=F_2sin(\phi_2)

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

\sum F_x=0\\F_{1x}=F_{2x}\\F_1cos(\phi_1)=F_2cos(\phi_2)\\\frac{F_1}{F_2}=\frac{cos(\phi_2)}{cos(\phi_1)}-------1

Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.

At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces F_{1y}\ and\ F_{2y} will be zero. This gives,

-F_{1y}\times x + F_{2y}(L-x) = 0\\F_{1y}\times x=F_{2y}(L-x)\\x=\frac{F_{2y}(L-x)}{F_{1y}}

But, F_{1y}=F_1sin(\phi_1)\ and\ F_{2y}=F_2sin(\phi_2)

Therefore,

x=\frac{F_2sin(\phi_2)(L-x)}{F_1sin(\phi_1)}\\\textrm{From equation (1),}\frac{F_2}{F_1}=\frac{cos(\phi_1)}{cos(\phi_2)}\\\therefore x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times (L-x)\\x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times L-\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times x\\\\

x(1+\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L\\x(1+\frac{cos(\phi_1)}{sin(\phi_1)}\times \frac{sin(\phi_2}{cos(\phi_2)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L

We know,

tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}

∴x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}

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2 years ago
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