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AysviL [449]
3 years ago
15

How much heat would be absorbed by 75.20 g of iron when heated from 22 C to 28 C

Physics
2 answers:
Bumek [7]3 years ago
8 0

Mass x SH x °C (or K) ΔT

= 75g x 0.45J/g/K x 6.0 ΔT

= 202.5 Joules of heat absorbed.


(202.5J / 4.184J/cal = 48.4 calories).

I guess that is the answer

____ [38]3 years ago
5 0

Hello friend ☺

ΔH = MCΔT

ΔH = to the amount of energy or change in energy (J)

M = mass of substance

C = waters specific heat capacity

ΔT = change in temperature

and so ΔH = 75.20 * 4.18 * ( 28-22 ) = 1886.016 J

Thanks ❤

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The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

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Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

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tia_tia [17]

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t = \frac{2}{-24cos\theta + 3}

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Ronch [10]

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The amplitude of the magnetic field at this surface is 1.56421\times 10^{-6}\ T

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