The answer would be number four. I'm sorry if I am too late. Byes.....
-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.
-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.
-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
</em> That's about <em>2.634 m/s²</em> <em>
</em>(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)
-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is
Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.
That's the force that's accelerating the little block, so that must be the tension
in the string.
Answer:
z = 3,737 10⁵ m
Explanation:
a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation
P V = n R T
P = (n r / V) T
We replace
P = (n R / V) T₀
b) Let's apply this equation in the points
Lower
.z = 0
P₀ = 610 Pa
P₀ = (nR / V) T₀
Higher.
P = 10 Pa
P = (n R / V) T₀ e^{- C z}
We replace
P = P₀ e^{- C z}
e^{- C z} = P / P₀
C z = ln P₀ / P
z = 1 / C ln P₀ / P
Let's calculate
z = 1 / 1.1 10⁻⁵ ln (610/10)
z = 3,737 10⁵ m