What is NOT one of the three primary resources that families have to reach financial goals? It is c) education
Answer:
Final volume, V2 = 24.62 L
Explanation:
Given the following data;
Initial volume = 40 L
Initial pressure = 80 Pa
Final pressure = 130 Pa
To find the final volume V2, we would use Boyles' law.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
Substituting into the equation, we have;
![80 * 40 = 130V_{2}](https://tex.z-dn.net/?f=%2080%20%2A%2040%20%3D%20130V_%7B2%7D%20)
![3200 = 130V_{2}](https://tex.z-dn.net/?f=%203200%20%3D%20130V_%7B2%7D%20)
![V_{2} = \frac {3200}{130}](https://tex.z-dn.net/?f=%20V_%7B2%7D%20%3D%20%5Cfrac%20%7B3200%7D%7B130%7D)
![V_{2} = 24.62](https://tex.z-dn.net/?f=%20V_%7B2%7D%20%3D%2024.62%20)
Final volume, V2 = 24.62 Liters
Answer:
D. infinitely extended in all directions
Explanation:
A semi infinite solid is infinitely extended in every direction. It has a single surface and can extend when heat is applied.
The body of a semi infinite solid is idealised, that is, when there is heat present, it expands in all directions to infinity. It can be used for a thick wall because its shape can be changed when subjected to different levels of heat near its surface.
It is also expands as heat is applied because its thickness is negligible.
This idealized body is used for earth, thick wall, steel piece of any shaped quenched rapidly etc indetermining variation of temperature near its surface & other surface being too far to have any impact on the region in short period of time since heat doesn’t have sufficient time to penetrate deep into body thus thickness can be neglected
Answer:
a) ![\Delta{t} = 5.39s](https://tex.z-dn.net/?f=%5CDelta%7Bt%7D%20%3D%205.39s)
b) the motorcycle travels 155 m
Explanation:
Let
, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):
![v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}](https://tex.z-dn.net/?f=v_%7Bm2%7D%3Dv_0%2Ba%5CDelta%7Bt%7D%5C%5Cx%2Bd%3D%28%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%20%29%5CDelta%7Bt%7D%5C%5Cv_c%20%3D%20v_0%20%3D%20%5Cfrac%7Bx%7D%7B%5CDelta%7Bt%7D%7D)
where:
is the speed of the motorcycle at time 2
is the velocity of the car (constant)
is the velocity of the car and the motorcycle at time 1
d is the distance between the car and the motorcycle at time 1
x is the distance traveled by the car between time 1 and time 2
Solving the system of equations:
![\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dcar%26motorcycle%5C%5Cx%3Dv_0%5CDelta%7Bt%7D%26x%2Bd%3D%28%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%7D%29%20%5CDelta%7Bt%7D%5Cend%7Barray%7D%5Cright%5D)
![v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s](https://tex.z-dn.net/?f=v_0%5CDelta%7Bt%7D%3D%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%5CDelta%7Bt%7D-d%20%5C%5C%5Cfrac%7Bv_0%2Bv_%7Bm2%7D%7D%7B2%7D%5CDelta%7Bt%7D-v_0%5CDelta%7Bt%7D%3Dd%5C%5C%28v_0%2Bv_%7Bm2%7D%29%5CDelta%7Bt%7D-2v_0%5CDelta%7Bt%7D%3D2d%5C%5C%28v_0%2Bv_0%2Ba%5CDelta%7Bt%7D%29%5CDelta%7Bt%7D-2v_0%5CDelta%7Bt%7D%3D2d%5C%5C%282v_0%2Ba%5CDelta%7Bt%7D%29%5CDelta%7Bt%7D-2v_0%5CDelta%7Bt%7D%3D2d%5C%5Ca%5CDelta%7Bt%7D%5E2%3D2d%5C%5C%5CDelta%7Bt%7D%3D%5Csqrt%7B%5Cfrac%7B2d%7D%7Ba%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%2A58%7D%7B4%7D%7D%3D%5Csqrt%7B29%7D%3D5.385s)
For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:
![x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933](https://tex.z-dn.net/?f=x%20%3D%20v_0%5CDelta%7Bt%7D%3D%2018%5Csqrt%7B29%7D%3D96.933m%5C%5Cthen%3A%5C%5Cx%2Bd%20%3D%20154.933)
Answer:
<u>The magnitude of the friction force is 8197.60 N</u>
Explanation:
Using the definition of the centripetal force we have:
![\Sigma F=ma_{c}=m\frac{v^{2}}{R}](https://tex.z-dn.net/?f=%5CSigma%20F%3Dma_%7Bc%7D%3Dm%5Cfrac%7Bv%5E%7B2%7D%7D%7BR%7D)
Where:
- m is the mass of the car
- v is the speed
- R is the radius of the curvature
Now, the force acting in the motion is just the friction force, so we have:
<u>Therefore the magnitude of the friction force is 8197.60 N</u>
I hope it helps you!