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nikdorinn [45]
3 years ago
10

A wall with a surface area of 10 m2 is 2.5 cm thick. The inner surface temperature of the wall is 4150C, but the outer surface i

s much cooler. The heat loss through the wall is 3 kW, and the thermal conductivity is 0.2 W/m-K. a. What is the temperature of the outer surface of the wall? List all applicable assumptio
Physics
1 answer:
klio [65]3 years ago
5 0

Answer:

650.65 K or 377.5°C

Explanation:

Area = A = 10 m²

Thickness of wall = L = 2.5 cm = 2.5×10⁻² m

Inner surface temperature of wall = T_i = 415°C = 688.15 K

Outer surface temperature of wall = T_o

Heat loss through the wall = 3 kW = 3×10³ W

Thermal conductivity of wall = k = 0.2 W/m K

Assumptions made here as follows

  1. There is not heat generation in the wall itself
  2. The heat conduction is one dimensional
  3. Heat flow follows steady state
  4. The material has same properties in all directions i.e., it is homogeneous.

Considering the above assumptions we use the following formula

Q=\frac {T_i-T_o}{\frac{L}{kA}}\\\Rightarrow T_o=T_i-\frac {QL}{kA}\\\Rightarrow T_o=688.15-\frac {3\times 10^{3}\times 2.5\times 10^{-2}}{0.2\times 10}\\\Rightarrow T_o=650.65~K

∴ The temperature of the outer surface of the wall is 650.65 K or 377.5°C

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Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

k = \frac{155}{0.10} N/m

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so now we have

2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}

m = 8.11 kg

so mass of the fish is 8.11 kg

4 0
3 years ago
A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha
Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

x(t) = A\cdot \cos (\omega \cdot t + \phi)

Where:

x(t) - Position of the mass with respect to the equilibrium position, measured in centimeters.

A - Amplitude of the mass-spring system, measured in centimeters.

\omega - Angular frequency, measured in radians per second.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)

Where the maximum acceleration of the system is represented by \omega^{2}\cdot A.

The natural frequency of the mass-spring system is:

\omega = \sqrt{\frac{k}{m} }

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

If k = 12\,\frac{N}{m} and m = 0.40\,kg, the natural frequency is:

\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }

\omega \approx 5.477\,\frac{rad}{s}

Lastly, the maximum acceleration of the system is:

a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)

a_{max} = 359.970\,\frac{cm}{s^{2}}

The maximum acceleration of the system is 359.970 centimeters per square second.

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MAVERICK [17]

Answer:

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Explanation:

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