Answer:
15 ohm
3 A
Explanation:
R =15 ohm (series = summation)
I= V/R = 45 / 15 = 3 A
The period is simply the inverse of the frequency, therefore:
T = 1 / f
T = 1 / 775 Hz
T = 0.001290 s → possible answer
T = 1.29 × 10⁻³ s → possible answer
T = 1.29 ms → possible answer
1. I think is letter b.
2. I thinks it's letter c.
Answer:
no sabo por q no mi entender
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
