An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet.
As a result, the object's <em>mass remains the same</em>, and its <em>weight decreases</em> to 1/4 of whatever it is when the object is on the planet's surface.
Answer : The final pressure in the two containers is, 2.62 atm
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

Thus, the expression for final pressure in the two containers will be:


where,
= pressure of N₂ gas = 4.45 atm
= pressure of Ar gas = 2.75 atm
= volume of N₂ gas = 3.00 L
= volume of Ar gas = 2.00 L
P = final pressure of gas = ?
V = final volume of gas = (4.45 + 2.75) L = 7.2 L
Now put all the given values in the above equation, we get:


Thus, the final pressure in the two containers is, 2.62 atm
Answer:
a) t = 1.6 s
b) d = 4.9 m
c) v = 16 m/s
d) θ = 79°
Explanation:
time of fall
t = √(2h/g) = √(2(12)/9.8) = 1.5649... s
d = vt = 3.1(1.56) = 4.8512...
vertical velocity vy = at = 9.8(1.56) = 15.336... m/s
v = √(15.336² + 3.1²) = 15.6464... m/s
θ = arctan(15.336/3.1) = 78.5724...°