Explanation:
cooking of duck will cost 48000
by the help of the method of rate × A + €
Find the magnitude of the steady-state response of the system whose system model is given by dx(t)/dt+ x(t)-f(t), where f(t) 2cos8t. Keep 3 significant figures
This question is incomplete, the complete question is;
Find the magnitude of the steady-state response of the system whose system model is given by
dx(t)/dt + x(t) = f(t)
where f(t) = 2cos8t. Keep 3 significant figures
Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )
Explanation:
Given that;
dx(t)/dt + x(t) = f(t) where f(t) = 2cos8t
dx(t)/dt + x(t) = f(t)
we apply Laplace transformation on both sides
SX(s) + x(s) = f(s)
(S + 1)x(s) = f(s)
f(s) / x(s) = S + 1
x(s) / f(s) = 1 / (S + 1)
Therefore
transfer function = H(s) = x(s)/f(s) = 1/(S+1)
f(t) = 2cos8t → [ 1 / ( S + 1 ) ] → x(t) = Acos(8t - ∅ )
A = Magnitude of steady state output
S = jw
S = j8
so
A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )
A = 2/√65 = 0.2481
∅ = tan⁻¹( 1/1) = 45°
therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric conditions, what is the maximum pressure on her hand? What would be the maximum pressure if the "car" were an Indy 500 racer traveling 200 mph?
Answer:
[tex]10.8\ \text{lb/ft^2}[/tex]
[tex]101.96\ \text{lb/ft}^2[/tex]
Explanation:
[tex]v_1[/tex] = Velocity of car = 65 mph = [tex]65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}[/tex]
[tex]\rho[/tex] = Density of air = [tex]0.00237\ \text{slug/ft}^3[/tex]
[tex]v_2=0[/tex]
[tex]P_1=0[/tex]
[tex]h_1=h_2[/tex]
From Bernoulli's law we have
[tex]P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}[/tex]
The maximum pressure on the girl's hand is [tex]10.8\ \text{lb/ft^2}[/tex]
Now [tex]v_1[/tex] = 200 mph = [tex]200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}[/tex]
[tex]P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2[/tex]
The maximum pressure on the girl's hand is [tex]101.96\ \text{lb/ft}^2[/tex]
Consider diodes in a rectifier circuit. Input voltage is sinusoidal with a peak of +/-10 V. Diode drop is 0.7 V. What is the PIV for each type rectifier 1. 0.7 V 2. 1.4 V 3. 10.7 V 4. 11.4 V Bridge rectifier 5. 19.3 V Full-wave rectifier 6. 8.6 V 7. 9.3 V Half-wave rectifier 8. 7.2 V 9. 12.1 V 10. 12.8 V 11. 10 V
Answer is given below:
Explanation:
Peak inverse voltage (PIV) can be defined as the maximum value of the reverse voltage of the diode, which is the maximum value of the input cycle when the diode is on. In reverse bias. Happens. 9.3V for braid rectifiers cut at 0.7The center tapered rectifier has 2 diodes in parallel so the maximum voltage is 2Vm so the answer to cut off the 0.7 voltage is19.3V. For a half wave rectifier it is Vm i.e. 10 V.Which type of forming operation produces a higher quality surface finish, better mechanical properties, and closer dimensional control of the finished piece?A. Hot working.B. Cold working.
Answer:
Option B (Cold working) would be the correct alternative.
Explanation:
Cold working highlights the importance of reinforcing material without any need for heat through modifying its structure or appearance. Metal becomes considered to have been treated in cold whether it is treated economically underneath the material's transition temperature. The bulk of cold operating operations are carried out at room temperature.The other possibility isn't linked to the given scenario. Therefore the alternative above is the right one.
Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 and 0.049, respectively? Why or why not?
Answer:
Yes it is possible.
Explanation:
This problem is about to possibility to have alloy of iron-carbon for which mass fraction of ferrite, [tex]$W_{\alpha} = 0.846$[/tex] and proeutectoid cementite, [tex]$W_{Fe_3C}=0.049$[/tex]
An alloy formation is possible when the composition values of the two alloy are equal.
Now writing the expression for the mass fraction of total ferrite, we have
[tex]$W_{\alpha}=\frac{C_{Fe_3C}-C_0}{C_{Fe_3C}-C_{\alpha}}$[/tex]
[tex]$0.846}=\frac{6.70-C_0}{6.70-0.022}$[/tex]
[tex]$5.649588 = 6.70 - C_0$[/tex]
[tex]$\therefore C_0 = 1.05 $[/tex] wt. % of C
Now write the expression for the mass fraction of the proeutectoid cementite :
[tex]$W_{Fe_3C}=\frac{C_1-0.76}{5.94}$[/tex]
[tex]$0.049=\frac{C_1-0.76}{5.94}$[/tex]
[tex]$C_1 = 1.05$[/tex] % wt. C
Since, [tex]$C_0 =C_1$[/tex], it is possible to have an alloy of iron - carbon.
Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
Sieve size Weight retained (g) No. 4 59.5 No. 8 86.5 No. 16 138.0 No. 30 127.8 No. 50 97.0 No. 100 66.8 Pan 6.3
Solution :
Sieve Size (in) Weight retain(g)
3 1.62
2 2.17
[tex]$1\frac{1}{2}$[/tex] 3.62
[tex]$\frac{3}{4}$[/tex] 2.27
[tex]$\frac{3}{8}$[/tex] 1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan 6.3 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]
[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]
x = 0.2634 mm
Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]
x = 1.6317 mm
[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]
Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]
[tex]$Cu= \frac{1.6317}{0.2643}$[/tex]
Cu = 6.17
Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]
[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm