During that final period of time,
his acceleration is
(9 m/s - 5 m/s) / (4 sec) = 1 m/s² .
Did you have a question to ask ?
Answer:
![\omega=0.37 [rad/s]](https://tex.z-dn.net/?f=%5Comega%3D0.37%20%5Brad%2Fs%5D)
Explanation:
We can use the conservation of the angular momentum.
Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.
So we will have:
Now, we just need to solve it for ω.
I hope it helps you!
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
Answer:
24.5987 cm
Explanation:
A = 1900 cm^2
Let r be the radius of disc.
The area of disc is given by
A = π r²
Where, π = 31.4
1900 = 3.14 x r²
r² = 605.095
r = 24.5987 cm
Explanation
(m) is measured in kilograms (kg)
<h2>(F) is measured in newtons (N)</h2>
<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>
