A real number that is not a natural

A compound wall consists of parallel layers of two different materials, 10 cm of brick and 2 cm of wood. If the difference of te

RSB [31]

The temperature difference between the wooden wall is 16⁰C, and the heat current per square metre of the wall is 7,325 W/m².

<h3>Temperature difference between the wooden wall</h3>

The temperature difference between the wooden wall is calculated as follows;

Let the brick wall = wall A
Let the wooden wall = wall B
Let the area of the walls = A

$\frac{dQ_A}{dt } = \frac{dQ_B}{dt} \\\\\frac{K_A \Delta T_A \times A}{L_A} = \frac{K_B \Delta T _B \times A}{L_B}\\\\\frac{K_A \Delta T_A }{L_A} = \frac{K_B \Delta T _B}{L_B}\\\\\frac{0.5 \times 20}{0.1} = \frac{0.125 \times \Delta T _B }{0.02} \\\\100 = 6.25\Delta T _B\\\\\Delta T _B = \frac{100}{6.25} \\\\\Delta T _B = 16 \ ^oC$

<h3 /><h3>Heat flowing in the walls</h3>

Q = KL(ΔT)

Q = 0.5 x 0.1 x (20 + 273)

Q = 14.65 W

<h3>Heat current per square meter of the Walls</h3>

QA = W/A

QA = (14.65)/A

Let the area of the wall = 10 cm x 2 cm = 0.1 m x 0.02 m = 0.002 m².

QA = 14.65/0.002

QA = 7,325 W/m²

Learn more about thermal conductivity here: brainly.com/question/11213835

4 0

3 years ago

Covalent bonds ___ conduct electricity well. <br><br> Always <br> Usually<br> Rarely<br> Never

mr_godi [17]

They rarely conduct electricity well.

4 0

3 years ago

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A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

How do you find pressure related to temperature

Ivenika [448]

Temperature and pressure are directly proportional to each other. If this helps,please pick me the best. Also,please say thanks.

8 0

3 years ago

10 Select the correct answer. What property of a wave remains unchanged when a wave enters a different medium? O A amplitude OB.

Mama L [17]

a and b ik thats the awser becuse imjust did it

8 0

3 years ago

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A real number that is not a natural​

A real number that is not a natural