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3 years ago

6

A phone cord is 6.48 m long. The cord has a mass of 0.253 kg. A transverse wave pulse is produced by plucking one end of the tau

t cord. The pulse makes four trips down and back along the cord in 0.877 s. What is the tension in the cord

1 answer:

monitta3 years ago

6 0

Answer:

136.27 N

Explanation:

From the question,

The velocity of the pulse is given as

v = L/t................. Equation 1

Where v = velocity of the pulse, L = total length of the pulse, t = time.

Given: L = 4×2×6.48 (as the pulse makes four tips down and back) = 51.84 m, t = 0.877 s.

substitute into equation 1

v = 51.84/0.877

v = 59.11 m/s.

Using,

But,

v = √(T/m')................... Equation 2

Where T tension in the cord, m' = mass per unit length of the cord.

make T the subject of the equation

T = v²m'.................... Equation 3

Given: v = 59.11 m/s, m' = 0.253/6.48 = 0.039 kg/m

Substitute into equation 3

T = 59.11²(0.039)

T = 136.27 N

Hence the tension in the cord = 136.27 N

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2 years ago

what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame

Hitman42 [59]

Answer:

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Explanation:

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perpendicular to the wall and aiming to the center of rotation.

This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
This force has the following general expression:

$F_{c} = m* \omega^{2} * r (2)$

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center of rotation (the radius of the circle), and m the mass of the

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Since Fc is actually Fn, we can replace the right side of (2) in (1), as

follows:

$F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)$

When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

$m* g = m* \mu_{smin} * \omega^{2} * r (4)$

(The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)

Cancelling the masses on both sides of (4), we get:

$g = \mu_{smin} * \omega^{2} * r (5)$

Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

$60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)$

Replacing by the givens in (5), we can solve for μsmín, as follows:

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2 years ago

Which statement describes the atomic number?<br>

Vadim26 [7]

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Explanation:

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2 years ago

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Answer:

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Explanation:

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$\Delta \theta = 0.066\,rev$

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$a_{n} = 2.554\,\frac{mm}{s^{2}}$

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$a = 3.293\,\frac{mm}{s^{2}}$

8 0

2 years ago

Enunciado: Una bola se lanza verticalmente de la parte superior de un edificio con una velocidad inicial de 25 m/s. La bola impa

bearhunter [10]

Responder:

Explicación:

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Usando la ecuación para obtener la altura de caída

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t es el tiempo = 7 segundos

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7 0

2 years ago