If we use the equation:
N2 + 3H2 --> 2NH3
Then
1 mol of Nitrogen required 3 moles of Hydrogen
x mols : 6.34mols
X = 6.34/3
X = 2.11 moles of Nitrogen are required.
Answer:
(a) ![v_g_i=1.08\frac{m}{s}](https://tex.z-dn.net/?f=v_g_i%3D1.08%5Cfrac%7Bm%7D%7Bs%7D)
(b) ![v_p_i=-0.3\frac{m}{s}](https://tex.z-dn.net/?f=v_p_i%3D-0.3%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
According to the law of conservation of momentum:
![\Delta p=0\\p_i=p_f](https://tex.z-dn.net/?f=%5CDelta%20p%3D0%5C%5Cp_i%3Dp_f)
Initially, the girl and the plank are at rest. So, relative to the ice, we have:
![0=m_gv_g_i+m_pv_p_i\\v_p_i=-\frac{m_gv_g_i}{m_p_i}(1)](https://tex.z-dn.net/?f=0%3Dm_gv_g_i%2Bm_pv_p_i%5C%5Cv_p_i%3D-%5Cfrac%7Bm_gv_g_i%7D%7Bm_p_i%7D%281%29)
(a) The velocity of the girl relative to the ice is:
![v_g_i=v_g_p+v_p_i(2)](https://tex.z-dn.net/?f=v_g_i%3Dv_g_p%2Bv_p_i%282%29)
Here,
is the velocity of the girl relative to the plank and
is the velocity of the plank relative to the ice.
Replacing (1) in (2):
![v_g_i=v_g_p-\frac{m_gv_g_i}{m_p_i}\\v_g_i+\frac{m_gv_g_i}{m_p_i}=v_g_p\\v_g_i(1+\frac{m_g}{m_p})=v_g_p\\v_g_i=\frac{v_g_p}{1+\frac{m_g}{m_p}}\\v_g_i=\frac{1.38\frac{m}{s}}{1+\frac{45kg}{159kg}}\\v_g_i=1.08\frac{m}{s}](https://tex.z-dn.net/?f=v_g_i%3Dv_g_p-%5Cfrac%7Bm_gv_g_i%7D%7Bm_p_i%7D%5C%5Cv_g_i%2B%5Cfrac%7Bm_gv_g_i%7D%7Bm_p_i%7D%3Dv_g_p%5C%5Cv_g_i%281%2B%5Cfrac%7Bm_g%7D%7Bm_p%7D%29%3Dv_g_p%5C%5Cv_g_i%3D%5Cfrac%7Bv_g_p%7D%7B1%2B%5Cfrac%7Bm_g%7D%7Bm_p%7D%7D%5C%5Cv_g_i%3D%5Cfrac%7B1.38%5Cfrac%7Bm%7D%7Bs%7D%7D%7B1%2B%5Cfrac%7B45kg%7D%7B159kg%7D%7D%5C%5Cv_g_i%3D1.08%5Cfrac%7Bm%7D%7Bs%7D)
(b) According to (2), the velocity of the plank relative to the surface of ice is:
![v_p_i=v_g_i-v_g_p\\v_p_i=1.08\frac{m}{s}-1.38\frac{m}{s}\\v_p_i=-0.3\frac{m}{s}](https://tex.z-dn.net/?f=v_p_i%3Dv_g_i-v_g_p%5C%5Cv_p_i%3D1.08%5Cfrac%7Bm%7D%7Bs%7D-1.38%5Cfrac%7Bm%7D%7Bs%7D%5C%5Cv_p_i%3D-0.3%5Cfrac%7Bm%7D%7Bs%7D)
The negative sing indicates that the plank is moving to the left.
The horizontal component is 2.0 cos(30°) = 1.732 m/s²
The vertical component is 2.0 sin(30°) = 1 m/s²
Answer:
The planets and moons.
Explanation:
Planets follow an elliptical path around the sun (kinda oval shaped). Moons do the same to planets.
Given Information:
Energy of laser pulses = E = 100 mJ = 100×10⁻³ Joules
Time = t = 1 ns = 1×10⁻⁹ seconds
Required Information:
Instantaneous power = P = ?
Answer:
![Instantaneous \: Power = 100 \: Mwatt](https://tex.z-dn.net/?f=Instantaneous%20%5C%3A%20Power%20%3D%20100%20%5C%3A%20Mwatt)
Explanation:
The instantaneous power is the power dissipated at any instant of time whereas the average power is the power dissipated over a given time interval.
The instantaneous laser power during each pulse is given by
![Instantaneous \: Power = \frac{E}{t}](https://tex.z-dn.net/?f=Instantaneous%20%5C%3A%20Power%20%3D%20%5Cfrac%7BE%7D%7Bt%7D)
Where E is the energy of the laser pulses and t is the time that each pulse lasts.
![Instantaneous \: Power = \frac{100\times10^{-3}}{1\times10^{-9}}](https://tex.z-dn.net/?f=Instantaneous%20%5C%3A%20Power%20%3D%20%5Cfrac%7B100%5Ctimes10%5E%7B-3%7D%7D%7B1%5Ctimes10%5E%7B-9%7D%7D)
![Instantaneous \: Power = 1\times10^{8} \: watt](https://tex.z-dn.net/?f=Instantaneous%20%5C%3A%20Power%20%3D%201%5Ctimes10%5E%7B8%7D%20%5C%3A%20watt)
or
![Instantaneous \: Power = 100 \: Mwatt](https://tex.z-dn.net/?f=Instantaneous%20%5C%3A%20Power%20%3D%20100%20%5C%3A%20Mwatt)
Therefore, the instantaneous power of each pulse is 100 Mwatt.