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rewona [7]
3 years ago
7

7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o

f water at 22.7oC. When the two substances reach thermal equilibrium, the final temperature is 24.2oC. What is the mass of the copper block?
Physics
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

37.34372 kg

Explanation:

m = Mass

\Delta T = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

Q=mc\Delta T

In this case the heat transfer will be equal

m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg

Mass of copper block is 37.34372 kg

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A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?​
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Answer:

<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg

k = 0.4

F = 200 N

<u>To </u><u>find </u><u>-</u><u> </u> acceleration

<u>Solution </u><u>-</u><u> </u>

F= kMA

200 = 0.4 * 20 * acceleration

200 = 8 * a

a = 8/200

a = 0.04 m s²

<h3>a = 0.04 m s²</h3>
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2 years ago
Under which conditions are particles in a medium said to be in phase with one another?
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Explanation:

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The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spri
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Which two components must a vector quantity
Lemur [1.5K]

Answer:

d. Direction and magnitude

Explanation:

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3 years ago
A non-ideal 12.2 V battery is connected across a resistor R. The internal resistance of the battery is 1.9Ohm. Calculate the pot
Brums [2.3K]

Answer:

R=100 Ohm, V=11.97 volts and I=0.12 amperes

R=10 Ohm, V=10.25 volts and I=1.20 amperes

R=2 Ohm, V=6.26 volts

Explanation:

The potential difference (voltage) of a battery with internal resistance is:

V=\xi-Ir (1)

with \xi the electromotive force (the voltage the batteries say to has) , I the current and r the internal resistance. By Ohm's law the current that passes through the resistor is:

I=\frac{V}{R} (2)

using (2) on (1):

V=\xi-\frac{V*r}{R}

solving for V:

V+\frac{V*r}{R}=\xi

V=\frac{\xi}{1+\frac{r}{R}} (3)

R=100 Ohm

V=\frac{12.2}{1+\frac{1.9}{100}}=11.97 V

R=10 Ohm

V=\frac{12.2}{1+\frac{1.9}{10}}=10.25 V

R=2 Ohm

V=\frac{12.2}{1+\frac{1.9}{2}}=6.26 V

Because we have now the values of I on the circuit (is the same through all the components because is a series circuit)

We use back substitution on (1) to find the current:

R=100 Ohm

I=\frac{V}{R}=\frac{11.97}{100}=0.12 A

R=10 Ohm

I=\frac{V}{R}=\frac{11.97}{10}=1.20 A

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3 years ago
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