From an energy balance, we can use this formula to solve for the angular speed of the chimney
ω^2 = 3g / h sin θ
Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s
The formula for radial acceleration is:
a = rω^2
So,
a = 53.2 (0.987) = 52.494 /s^2
The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887
The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2
If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°
Answer:
Explanation:
10. As a spaceship is moving toward Earth, an Earthling measures its length to be 325 m, while the captain on board radios that her spaceship's length is 1150 m. (c = 3.00 × 108 m/s) (a) How fast is the rocket moving relative to Earth? (b) What is the TOTAL energy of a 75.0-kg crewman as measured by (i) the captain in the rocket and (ii) the Earthling?
Use KE= 1/2mv^2
So...
50,000=(.5)(1,000)v^2
50,000=500 x v^2
Divide 500 on both sides
100 = v^2
Square root both sides to get rid of v^2
Therefore v = 10 m/s
A line on a graph has a downward slope
