Answer:
U = 9.1 m/s
Explanation:
from the question we are given the following
time (t) = 1.8 s
angle = 23 degrees
acceleration due to gravity (g) = 9.8 m/s^{2}
let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below
v = u + 0.5at
- The final velocity (v) is zero since the ball comes to rest
- The time (t) it takes to get to the maximum height would be half the time it is in the air, t = 0.5 x 1.8 = 0.9
therefore
0 = u - (0.5 x 9.8 x 0.9)
u = 7.9 m/s
for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore
U sin 60 = 7.9
U = 7.9 ÷ sin 60
U = 9.1 m/s
It wouldn't indicate the same source if the paint was from a different place. If the paint was the same brand and color it would be but if not then no
decreased 5 times
Explanation: if the force increases 5 times between them would decrease 5 times
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>