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:
The line charge density is 
Explanation:
Given that,
Diameter = 2.54 cm
Distance = 19.6 m
Potential difference = 115 kV
We need to calculate the line charge density
Using formula of potential difference
Where, r = radius
V = potential difference
Put the value into the formula
Hence, The line charge density is
Im pretty sure it’s a because it makes more sense you know?.
Answer:
B. changing by a constant amount each second
Explanation:
thats my answer
