Heat rises, and it is warmer at the equator, so I think warm air would rise at the equator and move towards the cooler poles.
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s
Answer: The final temperature is 470K
Explanation: Using the relation;
Q= ΔU +W
Given, n = 2mol
Initial temperature T1= 345K
Heat =Q= 2250J
Workdone=W=-870J(work is done on gas)
T2 =Final temperature =?
ΔU =3/2nR(T2-T1)
ΔU=3/2 × 2 ×8.314 (T2 - 345)
ΔU=24.942(T2-345)
Therefore Q = 24.942(T2-345)+ (-870)
2250=24.942(T2-345)+ (-870)
125.09=(T2-345)
T2 =470K
Therfore the final temperature is 470K
Answer:
5.31143691523 m/s²
Explanation:
m = Mass = 280 g
x = Displacement of spring = 21.7 cm
Time period

Angular velocity is given by


From Hooke's law

The acceleration due to gravity on the planet is 5.31143691523 m/s²
Yes, I have been able to satisfy my curiosity.