I believe the answer is C
Hope this helps :)
Answer:
Explanation:
We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.
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
<h2>
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>
Explanation:
One period means time taken to complete one revolution.
In case of swings in one period time it travels the same position through two times.
Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.
For 1 period = 2 times
For 3 periods = 3 x For 1 period
For 3 periods = 3 x 2 times
For 3 periods = 6 times
The child swing through the swing's equilibrium position 6 times during the course of 3 periods.
