His total displacement from his original position is -1 m
We know that total displacement of an object from a position x to a position x', d = final position - initial position.
d = x' - x
If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.
Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.
His final position, x" after moving back 6 m is gotten from
x" - x' = -6 m
x" = -6 + x'
x" = -6 + 5
x" = -1 m
Thus, his total displacement from his original position is
d = final position - initial position
d = x" - x
d = -1 m - 0 m
d = -1 m
So, his total displacement from his original position is -1 m
Learn more about displacement here:
brainly.com/question/17587058
I learned the equation as P•V=k•T .
So x=1, y=1, and z= -1 .
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
