Answer: (a) The magnitude of its temperature change in degrees Celsius is
.
(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is
.
Explanation:
(a) Expression for change in temperature is as follows.
![|\Delta T| = |x - y|K](https://tex.z-dn.net/?f=%7C%5CDelta%20T%7C%20%3D%20%7Cx%20-%20y%7CK)
= 15.1 K
= ![|(x - 273.15) - (y - 273.15)|^{o}C](https://tex.z-dn.net/?f=%7C%28x%20-%20273.15%29%20-%20%28y%20-%20273.15%29%7C%5E%7Bo%7DC)
= ![|x - y|^{o}C](https://tex.z-dn.net/?f=%7Cx%20-%20y%7C%5E%7Bo%7DC)
= ![15.1^{o}C](https://tex.z-dn.net/?f=15.1%5E%7Bo%7DC)
Therefore, the magnitude of its temperature change in degrees Celsius is
.
(b) Change in temperature from Celsius to Fahrenheit is as follows.
F = 1.8C + 32
C = ![\frac{F - 32}{1.8}](https://tex.z-dn.net/?f=%5Cfrac%7BF%20-%2032%7D%7B1.8%7D)
Since, K = C + 273
or, ![\Delta K = \Delta C = \frac{\Delta F}{1.8}](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20%5CDelta%20C%20%3D%20%5Cfrac%7B%5CDelta%20F%7D%7B1.8%7D)
![\Delta F = 1.8 \Delta K](https://tex.z-dn.net/?f=%5CDelta%20F%20%3D%201.8%20%5CDelta%20K)
= 1.8 (15.1)
= ![27.18^{o}F](https://tex.z-dn.net/?f=27.18%5E%7Bo%7DF)
or, = ![27.2^{o}F](https://tex.z-dn.net/?f=27.2%5E%7Bo%7DF)
Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is
.
Answer:
Explanation:
on volleyballit's clockwise and What Would Happen If Someone Served Out Of Order? If the wrong player on your team serves because it's not their turn, the point and the possession of the serve will go to your opponents.
Answer:what is the question exactly
Explanation:
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
![T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}](https://tex.z-dn.net/?f=T_2%20%3D%20T_1%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%5E%7B%5Cfrac%7Bk-1%20%7D%7Bk%7D)
k = 1.4
![T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K](https://tex.z-dn.net/?f=T_2%20%3D%20T_1%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%5E%7B%5Cfrac%7Bk-1%20%7D%7Bk%7D%7D%5C%5C%5C%5CT_2%20%3D%201200%28%5Cfrac%7B80%7D%7B150%7D%29%5E%7B%5Cfrac%7B1.4-1%20%7D%7B1.4%7D%7D%5C%5C%5C%5CT_2%20%3D%201002.714K)
Work done is given as;
![W = \frac{1}{2} *m*(v_i^2 - v_e^2)](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2Am%2A%28v_i%5E2%20-%20v_e%5E2%29)
inlet velocity is negligible;
![v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s](https://tex.z-dn.net/?f=v_e%20%3D%20%5Csqrt%7B%5Cfrac%7B2W%7D%7Bm%7D%20%7D%20%3D%20%5Csqrt%7B2%2AC_p%28T_1-T_2%29%7D%20%5C%5C%5C%5Cv_e%20%3D%20%5Csqrt%7B2%2A1004%281200-1002.714%29%7D%5C%5C%5C%5Cv_e%20%3D%20%5Csqrt%7B396150.288%7D%20%5C%5C%5C%5Cv_e%20%3D%20629.41%20%20%5C%20m%2Fs)
Therefore, the exit velocity is 629.41 m/s
Answer:
0.23 s
Explanation:
First of all, let's find the time constant of the circuit:
![\tau=RC](https://tex.z-dn.net/?f=%5Ctau%3DRC)
where
is the resistance
is the capacitance
Substituting,
![\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s](https://tex.z-dn.net/?f=%5Ctau%3D%2850%2C000%20%5COmega%29%282.0%5Ccdot%2010%5E%7B-6%7DF%29%3D0.1%20s)
The charge on a charging capacitor is given by
(1)
where
is the full charge
we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which
![Q(t)=0.90 Q_0](https://tex.z-dn.net/?f=Q%28t%29%3D0.90%20Q_0)
Substituting this into eq.(1) we find
![0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s](https://tex.z-dn.net/?f=0.90%20Q_0%20%3D%20Q_0%20%281-e%5E%7B-t%2F%5Ctau%7D%29%5C%5C0.90%3D1-e%5E%7B-t%2F%5Ctau%7D%5C%5Ce%5E%7B-t%2F%5Ctau%7D%3D1-0.90%3D0.10%5C%5C-%5Cfrac%7Bt%7D%7B%5Ctau%7D%3Dln%280.10%29%5C%5Ct%3D-%5Ctau%20ln%280.10%29%3D%280.1%20s%29ln%280.10%29%3D0.23%20s)