Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
![ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%20%3D%20%5Cfrac%7B%5Cdelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
![ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%20%5Cfrac%7B31%2C000%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B280.6%7D-%20%5Cfrac%7B1%7D%7B333.6%7D%5D%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%280.003564%20-%200.002998%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%20%280.000566%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%202.1104%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%20e%5E%7B2.1104%7D%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%208.2515%5C%5C%5C%5CP_2%20%3D%20%2840.1%2A8.2515%29mmHg%20%3D%20330.89%20mmHg)
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg
Answer:
15.825 m
Explanation:
t = Time taken = 2.5 s
u = Initial velocity = 6.75 m/s
v = Final velocity = 5.91 m/s
s = Displacement
a = Acceleration
Equation of motion
![v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20a%3D%5Cfrac%7Bv-u%7D%7Bt%7D%5C%5C%5CRightarrow%20a%3D%5Cfrac%7B5.91-6.75%7D%7B2.5%7D%5C%5C%5CRightarrow%20a%3D-0.336%5C%20m%2Fs%5E2)
![v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20s%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%5C%5C%5CRightarrow%20s%3D%5Cfrac%7B5.91%5E2-6.75%5E2%7D%7B2%5Ctimes%20-0.336%7D%5C%5C%5CRightarrow%20s%3D15.825%5C%20m)
The distance Rickey slides across the ground before touching the base is 15.825 m
<span>The contact force that acts on objects in a liquid or gas and allows objects to float is called </span>Buoyancy.
Complete question:
A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
Answer:
The current in the circuit 7 ms later is 0.2499 A
Explanation:
Given;
Ideal inductor, L = 45-mH
Resistor, R = 60-Ω
Ideal voltage supply, V = 15-V
Initial current at t = 0 seconds:
I₀ = V/R
I₀ = 15/60 = 0.25 A
Time constant, is given as:
T = L/R
T = (45 x 10⁻³) / (60)
T = 7.5 x 10⁻⁴ s
Change in current with respect to time, is given as;
![I(t) = I_o(1-e^{-\frac{t}{T}})](https://tex.z-dn.net/?f=I%28t%29%20%3D%20I_o%281-e%5E%7B-%5Cfrac%7Bt%7D%7BT%7D%7D%29)
Current in the circuit after 7 ms later:
t = 7 ms = 7 x 10⁻³ s
![I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A](https://tex.z-dn.net/?f=I%28t%29%20%3D%20I_o%281-e%5E%7B-%5Cfrac%7Bt%7D%7BT%7D%7D%29%5C%5C%5C%5CI%20%3D0.25%281-e%5E%7B-%5Cfrac%7B7%2A10%5E%7B-3%7D%7D%7B7.5%2A10%5E%7B-4%7D%7D%7D%29%5C%5C%5C%5CI%20%3D%200.25%280.9999%29%5C%5C%5C%5CI%20%3D%200.2499%20%5C%20A)
Therefore, the current in the circuit 7 ms later is 0.2499 A
Really, Gundy ? ! ?
The formula for the car's speed is given and discussed in the box. The formula is
v = √(2·g·μ·d)
Then they <em>tell</em> you that μ is 0.750 , and then they <em>tell</em> you that d = 52.9 m . Also, everybody knows that 'g' is gravity = 9.8 m/s² .
They also tell us that the mass of the car is 1,000 kg, and they tell us that it took 3.8 seconds to skid to a stop. But we already <em>have</em> all the numbers in the formula <em>without</em> knowing the car's mass or how long it took to stop. The police don't need to weigh the car, and nobody was there to measure how long the car took to stop. All they need is the length of the skid mark, which they can measure, and they'll know how fast the guy was going when he hit the brakes !
Now, can you take the numbers and plug them into the formula ? ! ?
v = √(2·g·μ·d)
v = √( 2 · 9.8 m/s² · 0.75 · 52.9 m)
v = √( 777.63 m²/s²)
v = 27.886 m/s
Rounded to 3 digits, that's <em>27.9 m/s </em>.
That's about 62.4 mile/hour .