Answer:
a. 51.84Kj
b. 2808.99 W/m^2
c. 11.75%
Explanation:
Amount of heat this resistor dissipates during a 24-hour period
= amount of power dissipated * time
= 0.6 * 24 = 14.4 Watt hour
(Note 3.6Watt hour = 1Kj )
=14.4*3.6 = 51.84Kj
Heat flux = amount of power dissipated/ surface area
surface area = area of the two circular end + area of the curve surface
= 2.136 *10^-4
Heat flux = = 2808.99
fraction of heat dissipated from the top and bottom surface
=11.75%
The <em>equivalent force-couple system</em> at O is the force and couple experienced when at point O due to the applied force at point A
The <em>equivalent force couple</em> system at O due to force <em>F</em> are;
Force, F = (<u>8.65·i - 4.6·j</u>) KN
Couple, M₀ ≈ <u>40.9 </u>k kN·m
The reason the above values are correct is as follows:
The known values for the <em>cantilever</em> are;
The <em>height </em>of the beam = 0.65 m
The <em>magnitude of </em>the applied <em>force</em>, F = 9.8 kN
The <em>length </em>of the beam = 4.9 m
The <em>angle </em>away from the vertical the force is applied = 26°
The required parameter:
The <em>equivalent force-couple system</em> at the centroid of the beam cross-section of the cantilever
Solution:
The <em>equivalent force-couple system</em> is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;
The <em>equivalent force</em> = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j
Which gives;
The <em>equivalent force</em> ≈ (<u>8.65·i - 4.6·j</u>) KN
The <em>couple </em><em>acting </em>at point O due to the force <em>F</em> is given as follows;
The <em>clockwise moment</em> = <em>9.8 kN × cos(28°) × 4.9</em>
The <em>anticlockwise moment</em> = <em>9.8 kN × sin(28°) 0.65/2 </em>
The sum of the moments = Anticlockwise moment - Clockwise moments
∴ The <em>sum </em>of the moments, ∑M, gives the moment acting at point O as follows;
M₀ = <em>9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9</em> ≈ 40.9 kN·m
The couple acting at O, due to F, M₀ ≈ <u>40.9 kN·m</u>
The equivalent force couple system acting at point O due the force, F, is as follows
F = (8.65·i - 4.6·j) KN
M₀ ≈ <u>40.9 </u>k kN·m
Learn more about equivalent force systems here:
Answer:
Vab ≈ 3.426 V
Explanation:
First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...
R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600
Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:
Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1
The voltage at Vb is also a divider from V1:
Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1
The parallel branches containing Va and Vb have an effective resistance of ...
(1030)(1710)/(1030+1710) = 642.81
That forms a divider with R1 to give V1:
V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V
The difference Va-Vb is ...
Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V
_____
We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...
(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0
(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0
(V1 -Va)/R3 -Va/R4 = 0
The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.
