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3 years ago

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A copper bar of length h and electric resistance r slides with negligible friction on metal rails that have negligible electric

resistance (see figure below. the rails are connected on the right with a wire of negligible electric resistance

1 answer:

Elanso [62]3 years ago

3 0

I think math is the best idea

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4 years ago

Read 2 more answers

The horizontal surface on which the block of mass 2.2 kg slides is frictionless. The force of 27 N acts on the block in a horizo

Digiron [165]

The magnitude of the resulting acceleration of the block is 6.14 m/s².

The given parameters;

<em>mass of the block, m = 2.2 kg</em>
<em>horizontal force on the block, F = 27 N</em>
<em>force below the horizontal, 81 N at 60⁰</em>

The net horizontal force on the block is calculated as follows;

$F_x = 27 \ - 81 \times cos (60)\\\\F_x = -13.5 \ N$

The acceleration of the block is calculated as follows;

$a = \frac{-13.5}{2.2} \\\\a = - 6.14 \ m/s^2\\\\a = 6.14 \ m/s^2 \ to \ the \ left$

Thus, the magnitude of the resulting acceleration of the block is 6.14 m/s².

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3 years ago

True [87]

Answer:

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

Explanation:

Given:

mass of ice, $m_i=14.7\ g=0.0147\ kg$

mass of water, $m_w=324\ g=0.324\ kg$

Assuming the initial temperature of the ice to be 0° C.

<u>Apply the conservation of energy:</u>

Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

<u>Now from the heat equation:</u>

$Q_i=Q_w$

$m_i.L=m_w.c_w.\Delta T$ ......................(1)

where:

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$c_w=$ specific heat of water $=4.186\ J.g^{-1}.^{\circ}C^{-1}$

$\Delta T=$ change in temperature

Putting values in eq. (1):

$14.7 \times 333.55=324\times 4.186\times \Delta T$

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

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