Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
Answer : 5m/s
Explanation:the formular for velocity is distance /time or you can say displacement /time. Then it would then be
100/20 =5m/s
Answer:
21870.3156 N
Explanation:
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 1.6 m/s²
Equation of motion

The acceleration of the craft should be 1.02234 m/s²

Weight of the craft

Thrust

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N
Answer
given,
For helium
Volume,V = 46 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₁ = ?
number of moles
we know
P V = n R T

n₁ = 1.89 moles
For oxygen
Volume,V = 12 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₂ = ?
number of moles
we know
P V = n R T

n₂ = 0.49 moles
Total volume of tank = 5 L
temperature of tank = 298 K
Partial pressure of helium


P₁ = 9.25 atm
Partial pressure of oxygen


P₂ = 2.39 atm
total pressure
P = P₁ + P₂
P = 9.25 + 2.39
P = 11.64 atm