Current will be
now just pluf in the values and Voila..
The gravitional potential energy, relative to the bottom of the giant drop, in joules, is (9800) times (the height of the drop in meters).
That's the PE of the empty car only, not counting any hapless screaming souls who may be trapped in it at that moment.
A is the answer you can believe me
Answer:
Explanation:
Let the radius of track required be r.
Centripetal force will be provided by frictional force which will be equal to
m v²/ r
Frictional force = mg x μ
So
m v² /r = mg μ
r = v² / μ g =
v = 29 km /h = 8.05 m /s
r =( 8.05 x 8.05 ) /( .32 x 9.8 ) = 20.66 m
Answer:
0.558 atm
Explanation:
We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT
Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:
P₀ = P₁ + P₂ + ....
P₀= total pressure
P₁=P₂= is the partial pressure of each gass
If we can consider that each gas is an ideal gas, then:
P₀= (nRT/V)₁ + (nRT/V)₂ +..
Considering the molecular mass of O₂:
M O₂= 32 g/mol
And also:
R= ideal gas constant= 0.082 Lt*atm/K*mol
T= 65°C=338 K
4.98 g O₂ = 0.156 moles O₂
V= 7.75 Lt
Then:
P°O₂=partial pressure of oxygen gas= (0.156x0.082x338)/7.75
P°O₂= 0.558 atm
