Answer:
10.93 rad/s
Explanation:
If we treat the student as a point mass, her moment of inertia at the rim is
![I_r = mr^2 = 67.8*3.7^2 = 928.182 kgm^2](https://tex.z-dn.net/?f=I_r%20%3D%20mr%5E2%20%3D%2067.8%2A3.7%5E2%20%3D%20928.182%20kgm%5E2)
So the system moment of inertia when she's at the rim is:
![I_1 = I_d + I_r = 274 + 928.182 = 1202.182 kgm^2](https://tex.z-dn.net/?f=I_1%20%3D%20I_d%20%2B%20I_r%20%3D%20274%20%2B%20928.182%20%3D%201202.182%20kgm%5E2)
Similarly, we can calculate the system moment of inertia when she's at 0.456 m from the center
![I_2 = I_d + 67.8*0.456^2 = 274 + 14.1 = 288.1 kgm^2](https://tex.z-dn.net/?f=I_2%20%3D%20I_d%20%2B%2067.8%2A0.456%5E2%20%3D%20274%20%2B%2014.1%20%3D%20288.1%20kgm%5E2)
We can apply the law of angular momentum conservation to calculate the post angular speed when she's 0.456m from the center:
![I_1\omega_1 = I_2\omega_2](https://tex.z-dn.net/?f=I_1%5Comega_1%20%3D%20I_2%5Comega_2)
![\omega_2 = \omega_1\frac{I_1}{I_2} = 2.62*\frac{1202.182}{288.1} = 10.93 rad/s](https://tex.z-dn.net/?f=%5Comega_2%20%3D%20%5Comega_1%5Cfrac%7BI_1%7D%7BI_2%7D%20%3D%202.62%2A%5Cfrac%7B1202.182%7D%7B288.1%7D%20%3D%2010.93%20rad%2Fs)
Answer:five times five is twenty five divded by 10 is 2.5 seconds of acceleration
Answer:
2.80N/m
Explanation:
Given data
mass m= 56kg
perios T= 11.2s
The expression for the period is given as
T=2π√m/k
Substitute
11.2= 2*3.142*√56/k
square both sides
11.2^2= 2*3.142*56/k
125.44= 351.904/k
k=351.904/125.44
k= 2.80N/m
Hence the spring constant is 2.80N/m
Here is the answer. What is happening at the atomic level to give rise to the observed energy is that t<span>he </span>atomic level<span> is affected by the movement of electrons so as to </span><span>give rise to the observed energy. Hope this answers your question. Have a great day!</span>
Answer:
Average speed is total distance divided by total time.
v = d / t