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2 years ago

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A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg

/m3. Note that this value is not the linear density of the cable. At what speed does a transverse wave move along the cable.

1 answer:

ElenaW [278]2 years ago

4 0

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

$Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3$

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

$p=m/V\\\\ p=m/(L*A)\\p*A=m/L$

Now the equation for speed of traverse wave is calculated through:

$\sqrt \frac{F*L}{m}\\$

= $\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}$

Substituting the values

$\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3} \\$

=22.49 m/s

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Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

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The angular velocity is $w_f = 1.531 \ rad/ s$

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$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

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$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

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2 years ago

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t

givi [52]

For t1:

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For t2:

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Wherein:
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Read 2 more answers

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love history [14]

Answer:

The energy is $9.4\times10^{-21}\ J$

(a) is correct option

Explanation:

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$Q=\Delta U+nRT$

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Put the value into the formula

$N=\dfrac{20}{28}\times6.02\times10^{23}$

$N=4.3\times10^{23}$

We need to calculate the energy

Using formula of energy

$E=\dfrac{\Delta U}{N}$

Put the value into the formula

$E=\dfrac{4022.73}{4.3\times10^{23}}$

$E=9.4\times10^{-21}\ J$

Hence, The energy is $9.4\times10^{-21}\ J$

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2 years ago