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3 years ago

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The masses of blocks A and B in the figure (Figure 1) are 20.0 kg and 10.0 kg, respectively. The blocks are initially at rest on

the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F⃗ is applied to the pulley.
Find the acceleration a⃗ A of block A when F is 124 N.

1 answer:

Anna35 [415]3 years ago

7 0

<span>You are given the masses of blocks A and B in the figure are 20.0 kg and 10.0 kg, respectively. The blocks are initially at rest on the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F⃗ is applied to the pulley. The acceleration is 4.13 m/s^2.

</span>

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A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

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Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

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$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

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c) The spring constant is:

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$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

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3 years ago

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Answer:

<em>Pure water </em><em>won’t conduct electricity</em>

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Thus absence of charge carriers make pure water non conducting. Charge carriers in metals are electrons and in ionic solutions they are positive and negative ions.

Tap water, aquarium water and ocean water contains dissolved electrolytes. When electricity is passed through them the electrolytes dissociates into ions and these ions conduct electricity.

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Answer:

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