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°
Is it not standard deviation? or am i just dumb lol
Answer:
The free body diagram is attached.
Explanation:
A force of 31[N] to the east, the second force goes to the south and it is equal to 28[N], the third force goes to the west and it is equal to 39 [N].
We can consider the crate as a particle. And all the forces are acting over the particle.
Vo = 18 m/s
angle 35 degrees
1) Components of the initial velocity
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s
2) Equations of postion:
x = Vox*t
y = Voy*t - gt^2 / 2
3) Calculations
A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s
x = 14.74 * t
t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m
t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m
t = 1.5s => x = 22.11 m
t = 2s => x = 29.48 m
B)
y = Voy*t - gt^2 / 2
Voy = 10.32 m/s
g = 10 m/s (approximation)
y = 10.32*t - 5t^2
t = 0.5 s=> y = 3.91m
t = 1 s => y = 5.32m
t = 1.5 s => y = 4.23m
t = 2 s => y = 0.64 m
screwdriver :)
hope you'll give a good review!
