Heeeelppp please um if your smart help

Answer:

The torque at the end of the beam is 2.5 Nm

Explanation:

Given;

length of beam, r = 0.5 m

applied force, F = 5 N

The torque at the end of the beam is given by;

τ = F x r

where;

τ  is the torque

F is applied force

r is length of the beam

τ = 5 x 0.5

τ = 2.5 Nm

Therefore, the torque at the end of the beam is 2.5 Nm

Answer:

the first one is the correct answer

Answer:

the first one would be correct

Explanation:

Answer:

The steady rate of heat transfer through the glass window is 707.317 watts.

Explanation:

A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

[tex]\dot Q_{total} = \dot Q_{cond} + \dot Q_{conv, in} + \dot Q_{conv, out}[/tex] (Eq. 1)

Given that window is represented as a flat element, we can expand (Eq. 1) as follows:

[tex]\dot Q_{total} = \frac{T_{i}-T_{o}}{R}[/tex] (Eq. 2)

Where:

[tex]T_{i}[/tex], [tex]T_{o}[/tex] - Indoor and outdoor temperatures, measured in Celsius.

[tex]R[/tex] - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

[tex]R = R_{cond} + R_{conv, in}+R_{conv, out}[/tex] (Eq. 3)

Where:

[tex]R_{cond}[/tex] - Conductive thermal resistance, measured in Celsius per watt.

[tex]R_{conv, in}[/tex], [tex]R_{conv, out}[/tex] - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

[tex]R = \frac{l}{k\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d} + \frac{1}{h_{i}\cdot w\cdot d}[/tex]

[tex]R = \frac{1}{w\cdot d}\cdot \left(\frac{l}{k}+\frac{1}{h_{i}}+\frac{1}{h_{o}} \right)[/tex] (Eq. 4)

Where:

[tex]w[/tex] - Width of the glass window, measured in meters.

[tex]d[/tex] - Length of the glass window, measured in meters.

[tex]l[/tex] - Thickness of the glass window, measured in meters.

[tex]k[/tex] - Thermal conductivity, measured in watts per meter-Celsius.

[tex]h_{i}[/tex], [tex]h_{o}[/tex] - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that [tex]w = 2.4\,m[/tex], [tex]d = 1.5\,m[/tex], [tex]l = 0.006\,m[/tex], [tex]k = 0.78\,\frac{W}{m\cdot ^{\circ}C}[/tex], [tex]h_{i} = 10\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex] and [tex]h_{o} = 25\,\frac{W}{m^{2}\cdot ^{\circ}C}[/tex], the overall thermal resistance is:

[tex]R = \left[\frac{1}{(2.4\,m)\cdot (1.5\,m)}\right] \cdot \left(\frac{0.006\,m}{0.78\,\frac{W}{m\cdot ^{\circ}C} }+\frac{1}{10\,\frac{W}{m^{2}\cdot ^{\circ}C} }+\frac{1}{25\,\frac{W}{m^{2}\cdot ^{\circ}C} } \right)[/tex]

[tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex]

Now, we obtain the steady rate of heat transfer from (Eq. 2): ([tex]R = 0.041\,\frac{^{\circ}C}{W}[/tex], [tex]T_{i} = -5\,^{\circ}C[/tex], [tex]T_{o} = 24\,^{\circ}C[/tex])

[tex]\dot Q_{total} = \frac{24\,^{\circ}C-(-5\,^{\circ}C)}{0.041\,\frac{^{\circ}C}{W} }[/tex]

[tex]\dot Q_{total} = 707.317\,W[/tex]

The steady rate of heat transfer through the glass window is 707.317 watts.