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mojhsa [17]
3 years ago
6

Energy is transferred as the heat between two objects, one with a temperature of 5 C and the other with a temperature of 20 C. I

f two other objects are to have the same amount of energy transferred between them, what might their temperatures be?
Physics
1 answer:
Delicious77 [7]3 years ago
8 0

Answer:

Final temperature is located between 5 °C and 20 °C.

Explanation:

The object with high temperature transfers energy to the object with low temperature until thermal equilibrium is reached, that is, when both objects have the same temperature.

Q_{H} = -Q_{L}

C_{H}\cdot (20^{\textdegree}C - T) = -C_{L}\cdot (5^{\textdegree}C-T)

20\cdot C_{H} - C_{H}\cdot T = -5\cdot C_{L}+ C_{L}\cdot T

(C_{H}+C_{L})\cdot T=20\cdot C_{H} + 5\cdot C_{L}

T = 20\cdot\frac{C_{H}}{C_{L}+C_{H}} + 5\cdot \frac{C_{L}}{C_{L}+C_{H}}

Now, let consider three scenarios:

Scenario 1 - C_{H} = C_{L]

T = 12.5^{\textdegree}C

Scenario 2 - C_{H}>> C_{L}

T = 20^{\textdegree}C

Scenario 3 - C_{H}

T = 5^{\textdegree}C

Final temperature is located between 5 °C and 20 °C.

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A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.550 m/s. The to
Arisa [49]

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Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

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V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

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v2 = 47.83 / 96.2

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3 0
3 years ago
A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
mr_godi [17]

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

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mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

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ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

5 0
3 years ago
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