Answer:
a)1.13×10³
b)1.6×10³
Explanation:
Given:
Boltzmann's constant (K)=1.38×10^-23 J/K
atmoic mass of helium = 4 AMU or 4×1.66×10^-27kg
a)The formula for RMS speed (Vrms) is given as
![Vrms=\sqrt{\frac{3KT}{m} }](https://tex.z-dn.net/?f=Vrms%3D%5Csqrt%7B%5Cfrac%7B3KT%7D%7Bm%7D%20%7D)
where
K= Boltzmann's constant
T= temperature
m=mass of the gas
![Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 206}{6.64\times 10^{-27}}}](https://tex.z-dn.net/?f=Vrms%3D%5Csqrt%7B%5Cfrac%7B3%5Ctimes%201.38%5Ctimes%2010%5E%7B-23%7D%5Ctimes%20206%7D%7B6.64%5Ctimes%2010%5E%7B-27%7D%7D%7D)
![Vrms=1.13\times 10^{3}m/s](https://tex.z-dn.net/?f=Vrms%3D1.13%5Ctimes%2010%5E%7B3%7Dm%2Fs)
b) RMS speed of helium when the temperature is doubled
![Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 2\times 206}{6.64\times 10^{-27}}}](https://tex.z-dn.net/?f=Vrms%3D%5Csqrt%7B%5Cfrac%7B3%5Ctimes%201.38%5Ctimes%2010%5E%7B-23%7D%5Ctimes%202%5Ctimes%20206%7D%7B6.64%5Ctimes%2010%5E%7B-27%7D%7D%7D)
![Vrms=1.598\times 10^{3}m/s](https://tex.z-dn.net/?f=Vrms%3D1.598%5Ctimes%2010%5E%7B3%7Dm%2Fs)
<span>Answer:
p[0] = 0
v[0] = 50 mi/h = 50 * 5280 / 3600 ft/s = 50 * (22/15) ft/s = 10 * 22/3 ft/s = 220/3 ft/s
a = -50 ft/s^2
a = -50
Integrate to find velocity
v = -50t + C
220/3 = -50 * 0 + C
220/3 = C
v = 220/3 - 50t
Find when v = 0
0 = 220/3 - 50 * t
50 * t = 220/3
t = 220 / 150
t = 22/15
Integrate the velocity function to find the position function
p = (220/3) * t - 25 * t^2 + C
0 = (220/3) * 0 - 25 * 0^2 + C
0 = C
p = (220/3) * t - 25 * t^2
t = 22/15
p = (220/3) * (22/15) - 25 * (484/225)
p = (4840 / 45) - 484 / 9
p = 484 * (10/45 - 1/9)
p = 484 * (2/9 - 1/9)
p = 484 * (1/9)
p = 484/9
p = 53.777777777777777777777777777778
53.8 feet</span>
Answer:
26.64 m
Explanation:
Given the following :
Acceleration at ocean surface = 0.0800 m/s²
Distance covered if initial speed = 0.700 m/s and accelerates to a speed of 2.18m/s
Using the equation :
v² = u² + 2as
Where ;
v = final velocity ; u = initial velocity ; a = acceleration ; s = distance covered
Therefore,
v² = u² + 2as
2.18² = 0.7² + (2 × 0.08 × s)
4.7524 = 0.49 + 0.16s
4.7524 - 0.49 = 0.16s
4.2624 = 0.16s
s = 4.2624 / 0.16
s = 26.64 m
Answer:
The mechanical energy of the satellite is about 12 Gigajoules
Explanation:
The total mechanical energy is the sum of the kinetic and potential energy:
![E = E_k+E_p\\E = \frac{1}{2}mv^2+ mgh](https://tex.z-dn.net/?f=E%20%3D%20E_k%2BE_p%5C%5CE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%2B%20mgh)
While we can determine the potential energy from the given values (height above earth's surface), to calculate the kinetic energy the velocity of the satellite needs to be determined first.
The formula for orbital velocity is:
![v = \sqrt {\frac{G\cdot m_E}{r}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%20%7B%5Cfrac%7BG%5Ccdot%20m_E%7D%7Br%7D%7D)
with G the gravitational constant, m_E the mass of the Earth, and r the satellite distance measured from the center of the Earth:
![v = \sqrt{\frac{6.673\cdot10^{-11} Nm^2/kg^2\cdot 5.98 \cdot 10^{24} kg}{6.38 \cdot 10^6 m+ 1.5\cdot 10^6m}}=7116.20 \frac{m}{s}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B6.673%5Ccdot10%5E%7B-11%7D%20Nm%5E2%2Fkg%5E2%5Ccdot%205.98%20%5Ccdot%2010%5E%7B24%7D%20kg%7D%7B6.38%20%5Ccdot%2010%5E6%20m%2B%201.5%5Ccdot%2010%5E6m%7D%7D%3D7116.20%20%5Cfrac%7Bm%7D%7Bs%7D)
This velocity points in the direction of the tangent of the orbit.
Now the kinetic and total mechanical energy can be calculated:
![E_k+E_p= \frac{1}{2}mv^2+ mgh=\\=\frac{1}{2}300kg\cdot 7116.2^2\frac{m^2}{s^2}+ 300kg\cdot9.8\frac{m}{s^2}\cdot 1.5\cdot 10^6m= \\= 12006045366J\approx12 GJ](https://tex.z-dn.net/?f=E_k%2BE_p%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%2B%20mgh%3D%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D300kg%5Ccdot%207116.2%5E2%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%2B%20300kg%5Ccdot9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%5Ccdot%201.5%5Ccdot%2010%5E6m%3D%20%5C%5C%3D%2012006045366J%5Capprox12%20GJ)
The mechanical energy of the satellite is about 12 Gigajoules
Rougher surfaces have more friction between them