A ball is thrown horizontally from a cliff at a speed of 10 m/s. you predict that its speed 1 s later will be slightly greater t
2 answers:
Answer:
Speed is slightly greater than 14 m/s.
Explanation:
The ball will under go projectile motion. The horizontal velocity remains constant while there would be increase in vertical velocity due to acceleration due to gravity in downward direction.
Using first equation of motion, after 1 s, vertical velocity will be:
v = u +at
v=0+(9.8 m/s²)(1 s) = 9.8 m/s
Horizontal velocity = 10 m/s
Net velocity:
Speed is slightly greater than 14 m/s.
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Answer:
<em>a. 4.21 moles</em>
<em>b. 478.6 m/s</em>
<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-
Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = <em>4.21 moles</em>
The equation for root mean square velocity is
Vrms =
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms = = <em>478.6 m/s</em>
<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>
=
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
=
= 0.66
Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>
<em>the root mean square velocity of the oxygen gas is </em>
<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>