Answer:
120 m/s
Explanation:
Given:
v₀ = 0 m/s
a = 12 m/s²
t = 10 s
Find: v
v = at + v₀
v = (12 m/s²) (10 s) + 0 m/s
v = 120 m/s
The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case,
kQ1/(r1)^2 = kQ2/(r2)^2 r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2 kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 = 1/4
Answer:
The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Explanation:
Given the data in the question;
Using the Clapeyron equation
![(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}](https://tex.z-dn.net/?f=%28%5Cfrac%7BdP%7D%7BdT%7D%20%29_%7Bsat%20%7D%20%3D%20%5Cfrac%7Bh_%7Bfg%7D%7D%7BTv_%7Bfg%7D%7D)
![(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }](https://tex.z-dn.net/?f=%28%5Cfrac%7BdP%7D%7BdT%7D%20%29_%7Bsat%20%7D%20%3D%20%5Cfrac%7B%5Cfrac%7BH_%7Bfg%7D%7D%7Bm%7D%20%7D%7BT%5Cfrac%7BV_%7Bfg%7D%7D%7Bm%7D%20%7D)
where
is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu
T is the temperature ( 15 + 460 )R
m is the mass of water ( 0.5 Ibm )
is specific volume ( 1.5 ft³ )
we substitute
/
272.98 Ibf-ft²/R
Now,
![(\frac{P_2 - P_1}{T_2 - T_1})_{sat](https://tex.z-dn.net/?f=%28%5Cfrac%7BP_2%20-%20P_1%7D%7BT_2%20-%20T_1%7D%29_%7Bsat)
where P₁ is the initial pressure ( 50 psia )
P₂ is the final pressure ( 60 psia )
T₁ is the initial temperature ( 15 + 460 )R
T₂ is the final temperature = ?
we substitute;
![= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}](https://tex.z-dn.net/?f=%3D%20%28%2015%20%2B%20460%20%29%20%2B%20%5Cfrac%7B%2860-50%29psia%28%5Cfrac%7B144in%5E2%7D%7Bft%5E2%7D%29%20%7D%7B272.98%7D)
![T_2 = 475 + 5.2751\\](https://tex.z-dn.net/?f=T_2%20%3D%20475%20%2B%205.2751%5C%5C)
480.275 R
Therefore, boiling point temperature of this substance when its pressure is 60 psia is 480.275 R
Answer:
You would be watching tv for 1 hour and 40 mins
Explanation:
How much time take it take for 3:50 am to 5:30