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3 years ago

14

A rope breaks when the tension reaches 205 N. What is the maximum speed at which it can swing a 0.477 kg mass in a circle of rad

ius 0.725 m?

1 answer:

jekas [21]3 years ago

8 0

Answer:

Maximum velocity will be 17.651 m /sec

Explanation:

We have given a rope breaks when tension reaches 205 N

Mass m = 0.477 kg

Radius of the circle r = 0.725 m

Tension in the rope is given by

$T=\frac{mv^2}{r}$

$205=\frac{0.477\times v^2}{0.725}$

$0.477v^2=148.625$

$v^2=311.5828$

v = 17.651 m /sec

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Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

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$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

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$P=\rho g h$

Put the value into the formula

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Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

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3 years ago

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Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

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Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

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Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

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3 years ago