Answer:
Yes
Explanation:
Because if you push it, the skateboard has kinetic/potential energy.
Answer:
Angular velocity is same as frequency of oscillation in this case.
ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Explanation:
- write the equation F(r) = -K
with angular momentum <em>L</em>
- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.
- Write the energy of the orbit in relative to r = 0, and solve for "E".
- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.
- Solve for effective potential
- ω =
x ![[\frac{L^{2}}{mK}]^{3/14}](https://tex.z-dn.net/?f=%5B%5Cfrac%7BL%5E%7B2%7D%7D%7BmK%7D%5D%5E%7B3%2F14%7D)
Answer:

Explanation:
Given that
The speed of the airplane ,v= 142 m/s
The speed of the air ,u = 30 m/s
Lets take angle make by airplane from east direction towards north direction is θ .
Now by using diagram ,we can say that

Now by putting the values in the above equation we get



Therefore the angle will be 12.19° .
The angular speed of the device is 1.03 rad/s.
<h3>What is the conservation of angular momentum?</h3>
A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.
Using the conservation of angular momentum

Here, = the system's angular momentum before the collision
= 0 + mv
= (0.005)(450)(0.752)
= 1.692 kgm²/s
The moment of inertia of the system is given by
I = 2(M₁R₁² + M₂R₂²)+ mR₁²
= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²
= 1.6292 kgm²
Here, = Iω
So,
1.692 = 1.6292(ω)
ω = 1.03 rad/s
To know more about the conservation of angular momentum, visit:
brainly.com/question/1597483
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