Explanation:
From Newton's second law:
F = ma
Given that m = 4 kg and a = 8 m/s²:
F = (4 kg) (8 m/s²)
F = 32 N
If m is reduced to 1 kg and F stays at 32 N:
32 N = (1 kg) a
a = 32 m/s²
So the acceleration increases by a factor of 4.
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
A front is a narrow region between two air masses of different densities.
Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)
Finial speed= 27.8s
Initial speed= 0s
Time taken= 5.15
So..
27.8-0/5.15= 5.40m/s (rounded to two decimal places)