Answer: A)30V. First find the current of the circuit. I=V/R(total resistance). So I=60/120=0.5. Now to find voltage drop in R3 use ohms law as given. V(of 3)=(0.5)(60)=30V
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
The maximum speed is 10.4 m/s
Explanation:
For a body in uniform circular motion, the centripetal acceleration is given by:
![a=\frac{v^2}{r}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
v is the linear speed
r is the radius of the circular path
In this problem, we have the following data:
- The maximum centripetal acceleration must be
![a=1.1 g](https://tex.z-dn.net/?f=a%3D1.1%20g)
where
is the acceleration of gravity. Substituting,
![a=(1.1)(9.8)=10.8 m/s^2](https://tex.z-dn.net/?f=a%3D%281.1%29%289.8%29%3D10.8%20m%2Fs%5E2)
- The radius of the turn is
r = 10 m
Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:
![v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bar%7D%3D%5Csqrt%7B%2810.8%29%2810%29%7D%3D10.4%20m%2Fs)
Learn more about centripetal acceleration:
brainly.com/question/2562955
#LearnwithBrainly
Answer: I would choose options C and D
Explanation: