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nalin [4]
2 years ago
5

1. Determina el momento que produce una fuerza de 7 N tangente a una rueda de un metro de diámetro, sabiendo que el punto de apl

icación es el mismo borde de dicha rueda provocando un impulso en el sentido de las agujas del reloj. Solución:
Physics
1 answer:
Rudik [331]2 years ago
5 0

Answer:

τ= F r     into the blade

Explanation:

The moment of a force is defined by

         τ = F x r

where the bold indicates vectors

Let us write in the expression in magnitude

         τ = F r sin θ

in our case the force is tangent to the wheel therefore the angle between F and the radius is 90º, and the sin 90 = 1

       τ= F r

The direction of τ can be used by the rule of the right hand, the fingers curve in the direction of the torque when advancing from the force to the radius and the thumb points in the direction of the torque.

In this case, for a clockwise rotation, the fingers are curved in the direction and the thumb points into the blade, this is the direction of the τ.

TRASLATE

El momento de una fura es definido por

         τ = F x r

donde la negrillas indican vectores

Escribamos en ta expresión en magnitud

          τ = F r sin θ

en nuestro caso la fuerza es tangente a la rueda por lo tanto el angulo entre F y el radios es 90º, y el sin 90=1

        τ = F r

la dirección de tau la podemos  usar la regla de la mano derecha, los dedos curva en la dirección del torque al avanzar dese la fuerza al radio y el pulgar apunta en la dirección del torque.

En este caso para un giro en sentido horario los dedos se curvan ente sentido y el pulgar apunta hacia dentro de lla hoja, esta es la dirección del troque

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Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

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​

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

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6 0
2 years ago
An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. in
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A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar
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Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

\displaystyle V=k\frac{Q}{r}

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

d=\sqrt2 a

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}

\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m

The total potential is  

V_t=V_1+V_2+V_3+V_4

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of \pm 3\mu\ C. Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}

V_1=V_2=50912\ V

The total potential is

V_t=50912\ V+50912\ V=1\times 10^5\ V

\boxed{V_t=1\times 10^5\ V}

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8 0
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Yuki888 [10]

Answer:

The system is still balanced

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(3)T=(0)*1/2=0

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6 0
2 years ago
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