Answer:
81.6 m
Explanation:
Answer: 81.6 m.
The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is
-v(initial) / -g = t
-40 m/s / -9.8 m/s^2 = 4.08 s
The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.
So the distance d of maximum height is
d = v(avg)•t
d = 20 m/s • 4.08 s = 81.6 m.
Hi,
The force that acts on hydraulic machine is heavy therefore the content must be something that cannot be compressed by that kind of force, the gas can easily be compressed while a liquid is nearly impossible to.
Answer:
a) ![k_{avg}=6.22\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D6.22%5Ctimes%2010%5E%7B-21%7D)
b) ![k_{avg}=8.61\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D8.61%5Ctimes%2010%5E%7B-21%7D)
c) ![k_{mol}=3.74\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D3.74%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
d) ![k_{mol}=5.1\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D5.1%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
Explanation:
Average translation kinetic energy (
) is given as
....................(1)
where,
k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K
T = Temperature in kelvin
a) at T = 27.8° C
or
T = 27.8 + 273 = 300.8 K
substituting the value of temperature in the equation (1)
we have
![k_{avg}=6.22\times 10^{-21}J](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D6.22%5Ctimes%2010%5E%7B-21%7DJ)
b) at T = 143° C
or
T = 143 + 273 = 416 K
substituting the value of temperature in the equation (1)
we have
![k_{avg}=8.61\times 10^{-21}J](https://tex.z-dn.net/?f=k_%7Bavg%7D%3D8.61%5Ctimes%2010%5E%7B-21%7DJ)
c ) The translational kinetic energy per mole of an ideal gas is given as:
![k_{mol}=A_{v}\times k_{avg}](https://tex.z-dn.net/?f=k_%7Bmol%7D%3DA_%7Bv%7D%5Ctimes%20k_%7Bavg%7D)
here
= Avagadro's number; ( 6.02×10²³ )
now at T = 27.8° C
![k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D6.02%5Ctimes%2010%5E%7B23%7D%5Ctimes%206.22%5Ctimes%2010%5E%7B-21%7D)
![k_{mol}=3.74\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D3.74%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
d) now at T = 143° C
![k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D6.02%5Ctimes%2010%5E%7B23%7D%5Ctimes%208.61%5Ctimes%2010%5E%7B-21%7D)
![k_{mol}=5.1\times 10^{3}J/mol](https://tex.z-dn.net/?f=k_%7Bmol%7D%3D5.1%5Ctimes%2010%5E%7B3%7DJ%2Fmol)
1.8461 km/hr Well i need more characters so i might as well type a beautiful sentence for you to read and waste your time on.