Explanation:
a chip on your shoulder is an example
Answer:
Explanation:
it take oxygen in the atmosphere to burn it... in space there isn't any air :0
Answer:
I would say its a deep ocean trench
Explanation:
This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim
Answer:
![W_f = 2.319 rad/s](https://tex.z-dn.net/?f=W_f%20%3D%202.319%20rad%2Fs)
Explanation:
For answer this we will use the law of the conservation of the angular momentum.
![L_i = L_f](https://tex.z-dn.net/?f=L_i%20%3D%20L_f)
so:
![I_mW_m = I_sW_f](https://tex.z-dn.net/?f=I_mW_m%20%3D%20I_sW_f)
where
is the moment of inertia of the merry-go-round,
is the initial angular velocity of the merry-go-round,
is the moment of inertia of the merry-go-round and the child together and
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = ![\frac{1}{2}M_mR^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DM_mR%5E2)
I = ![\frac{1}{2}(115 kg)(2.5m)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28115%20kg%29%282.5m%29%5E2)
I = 359.375 kg*m^2
Where
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:
![I_s = \frac{1}{2}M_mR^2+mR^2](https://tex.z-dn.net/?f=I_s%20%3D%20%5Cfrac%7B1%7D%7B2%7DM_mR%5E2%2BmR%5E2)
![I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2](https://tex.z-dn.net/?f=I_s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28115kg%29%282.5m%29%5E2%2B%2823.5kg%29%282.5m%29%5E2)
![I_s = 506.25kg*m^2](https://tex.z-dn.net/?f=I_s%20%3D%20506.25kg%2Am%5E2)
Finally we replace all the data:
![(359.375)(3.2672) = (506.25)W_f](https://tex.z-dn.net/?f=%28359.375%29%283.2672%29%20%3D%20%28506.25%29W_f)
Solving for
:
![W_f = 2.319 rad/s](https://tex.z-dn.net/?f=W_f%20%3D%202.319%20rad%2Fs)
Answer:
The velocity of the frozen rock at
is -14.711 meters per second.
Explanation:
The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity (
), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:
(Eq. 1)
Where:
- Initial velocity, measured in meters per second.
- Gravity acceleration, measured in meters per square second.
- Time, measured in seconds.
If we get that
,
and
, then final velocity is:
![v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (1.5\,s)](https://tex.z-dn.net/?f=v%20%3D%200%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%2B%5Cleft%28-9.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%20%5Ccdot%20%281.5%5C%2Cs%29)
![v = -14.711\,\frac{m}{s}](https://tex.z-dn.net/?f=v%20%3D%20-14.711%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
The velocity of the frozen rock at
is -14.711 meters per second.